332354 + Interview Questions in Secondary School in Math

1.	Pls answer this question, forr the answer i will make it brilliant
Answer» Answer: 4/4 + 4/4 1+1= 2 MAY be CORRECT

1.

Pls answer this question, forr the answer i will make it brilliant

Answer»

Answer:

4/4 + 4/4

1+1= 2

MAY be CORRECT

Discussion

2.	What is an rational number
Answer» Answer: rational no. is the no. that can be expressed or fraction p/q of two integer

2.

What is an rational number

Answer»

Answer:

rational no.

is the no. that can be expressed or fraction p/q of two integer

Discussion

3.	If the system of the equations x-ky-z=0, kx-y-z=0, x+y-z=0 has a non-zero solution, then the possible values of k are:
Answer» ANSWER: $\huge\mathbb\purple{Q?=✓!}$

3.

If the system of the equations x-ky-z=0, kx-y-z=0, x+y-z=0 has a non-zero solution, then the possible values of k are:

Answer»

ANSWER:

$\huge\mathbb\purple{Q?=✓!}$

Discussion

4.	Find the least number to be subtracted from 2413 to make it a perfect square.
Answer» Answer: The least number to be SUBTRACTED from 2413 to make it a perfect SQUARE. is 12. Use DIVISION method to FIND the square root.

4.

Find the least number to be subtracted from 2413 to make it a perfect square.

Answer»

Answer:

The least number to be SUBTRACTED from 2413 to make it a perfect SQUARE. is 12.

Use DIVISION method to FIND the square root.

Discussion

5.	What will be the decimal expansion of 131/16,terminating or non terminating ?solve it to say
Answer» Step-by-step explanation: 131/16=8.1875 so we can OBSERVE that this is TERMINATING and non-repeating in NATURE

5.

What will be the decimal expansion of 131/16,terminating or non terminating ?solve it to say

Answer»

Step-by-step explanation:

131/16=8.1875

so we can OBSERVE that this is TERMINATING and non-repeating in NATURE

Discussion

6.	Plz plz plz plz answer it... answer the correct plz...plz..
Answer» ANSWER: PLZ FOLLOW me and MARK me as BRAINLIEST

6.

Plz plz plz plz answer it... answer the correct plz...plz..

Answer»

ANSWER:

PLZ FOLLOW me

and MARK me as BRAINLIEST

Discussion

7.	A contractor plans to install two slides for the children to play in a park . For the children below the age of 5years, she prefers to have a slide whose top is at a height of 1.5 metre . and is inclined at an angle of 30° to the ground , whereas for elder children , she. wants to have a steep slide at a height of 3metre , and inclined at an angle of 60° to the ground. what should be the length of the slide in each case?
Answer» Step-by-step explanation: YOUNGER children In △ABC sin30 o = AC BC 2 1 = AC 1.5 ∴AC=3m Elder children XZ YZ = 2 3 = XZ 3 XZ= 3 6 =2 3 m

7.

A contractor plans to install two slides for the children to play in a park . For the children below the age of 5years, she prefers to have a slide whose top is at a height of 1.5 metre . and is inclined at an angle of 30° to the ground , whereas for elder children , she. wants to have a steep slide at a height of 3metre , and inclined at an angle of 60° to the ground. what should be the length of the slide in each case?

Answer»

Step-by-step explanation:

YOUNGER children

In △ABC

sin30

o

=

AC

BC

2

1

=

AC

1.5

∴AC=3m

Elder children

XZ

YZ

=

2

3

=

XZ

3

XZ=

3

6

=2

3

m

Discussion

8.	Find the perimeter of a square San side is 5v2 cm
Answer» ANSWER: Square Solve for perimeter P = 20 a Side

8.

Find the perimeter of a square San side is 5v2 cm

Answer»

ANSWER:

Square

Solve for

perimeter

P

=

20

a

Side

Discussion

9.	Suman reads a book interrogative sentence
Answer» $\huge\mathbb\purple{ANSWER}$

9.

Suman reads a book interrogative sentence

Answer»

$\huge\mathbb\purple{ANSWER}$

Discussion

10.	In the given figure, PQR is an equilateraltriangle of side 20 cm. AQSR is inscribed in it,LQSR -90, Q5 - 16 cm. Find (1) SP, (ii) thearea of the shaded portion. [Take 13 -1.732).
Answer» ANSWER: Where is FIGURE! ..........

10.

In the given figure, PQR is an equilateraltriangle of side 20 cm. AQSR is inscribed in it,LQSR -90, Q5 - 16 cm. Find (1) SP, (ii) thearea of the shaded portion. [Take 13 -1.732).

Answer»

ANSWER:

Where is FIGURE!

..........

Discussion

11.	12. In a class of 100 students, 45students read Physics, 52 studentsread Chemistry and 15 students readboth the subjects. Find the numberof students who study neitherPhysics nor ChemistryO 16O 17O 18O 19t
Answer» 18 is the CORRECT answer. HOPE it helps U !!♡

11.

12. In a class of 100 students, 45students read Physics, 52 studentsread Chemistry and 15 students readboth the subjects. Find the numberof students who study neitherPhysics nor ChemistryO 16O 17O 18O 19t

Answer»

18 is the CORRECT answer.

HOPE it helps U !!♡

Discussion

12.	The monthly expenses of a student have increased from ₹350 to ₹500. find the ratioI) increase in expenses to original expenses ii) original expenses to incresed expenses iii) increased expenses to increase in expenses
Answer» i) 3:7 ii) 7:10 iii) 10:3 pls MARK me BRAINLIEST and FOLLOW me

12.

The monthly expenses of a student have increased from ₹350 to ₹500. find the ratioI) increase in expenses to original expenses ii) original expenses to incresed expenses iii) increased expenses to increase in expenses

Answer»

i) 3:7

ii) 7:10

iii) 10:3

pls MARK me BRAINLIEST and FOLLOW me

Discussion

13.	The molecular weight of Al2O3 if atomic mass of Al=27U and O=16U
Answer» ANSWER: xxxxxxxxq FU aiqisisisisjissiaihs KELSEY

13.

The molecular weight of Al2O3 if atomic mass of Al=27U and O=16U

Answer»

ANSWER:

xxxxxxxxq FU aiqisisisisjissiaihs KELSEY

Discussion

14.	6. ABC is a right-angled triangle. Angle ABC = 90°AC = 25 cm and AB - 24 cm. Calculate the area of aA.ABC
Answer» Answer: The AREA of the triangle is 84 cm². Step-by-step-explanation: NOTE: *Refer to the attachment for the diagram.* In figure, △ABC is a right-angled triangle. m∟ABC = 90° AC = 25 cm - - - [ Given ] AB = 24 cm We have to find the area of the triangle. In △ABC, m∟ABC = 90° ∴ ( AC )² = ( AB )² + ( BC )² - - [ Pythagoras theorem ] ⇒ ( 25 )² = ( 24 )² + ( BC )² ⇒ 625 = 576 + BC² ⇒ BC² = 625 - 576 ⇒ BC² = 49 ⇒ BC = √49 - - [ Taking SQUARE roots ] ⇒ BC = 7 cm Now, we know that, *Area of right-angled triangle = ½ Base * Height** ⇒ A ( △ABC ) = ½ * BC * AB ⇒ A ( △ABC ) = ½ * 7 * 24 ⇒ A ( △ABC ) = 7 * 12 ⇒ A ( △ABC ) = 84 cm² ∴ The area of the triangle is 84 cm². ───────────────────── Alternative Method: In △ABC, AB ( s₁ ) = 24 cm BC ( s₂ ) = 7 cm AC ( s₃ ) = 25 cm Now, we know that, Semi perimeter of triangle = ( Sum of sides ) / 2 ⇒ s ( △ABC ) = ( s₁ + s₂ + s₃ ) / 2 ⇒ s ( △ABC ) = ( 24 + 7 + 25 ) / 2 ⇒ s ( △ABC ) = 56 ÷ 2 ⇒ s ( △ABC ) = 28 cm Now, we know that, Area of triangle = √[ s ( s - s₁ ) ( s - s₂ ) ( s - s₃ ) ] ⇒ A ( △ABC ) = √[ 28 ( 28 - 24 ) ( 28 - 7 ) ( 28 - 25 ) ] ⇒ A ( △ABC ) = √( 28 * 4 * 21 * 3 ) ⇒ A ( △ABC ) = √( 112 * 63 ) ⇒ A ( △ABC ) = √[ ( 16 * 7 ) * ( 9 * 7 ) ] ⇒ A ( △ABC ) = √( 16 * 9 * 7 * 7 ) ⇒ A ( △ABC ) = √( 4 * 4 * 3 * 3 * 7 * 7 ) ⇒ A ( △ABC ) = 4 * 3 * 7 ⇒ A ( △ABC ) = 12 * 7 ⇒ A ( △ABC ) = 84 cm² ∴ The area of the triangle is 84 cm².

14.

6. ABC is a right-angled triangle. Angle ABC = 90°AC = 25 cm and AB - 24 cm. Calculate the area of aA.ABC

Answer»

Answer:

The AREA of the triangle is 84 cm².

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

In figure, △ABC is a right-angled triangle.

m∟ABC = 90°

AC = 25 cm - - - [ Given ]

AB = 24 cm

We have to find the area of the triangle.

In △ABC, m∟ABC = 90°

∴ ( AC )² = ( AB )² + ( BC )² - - [ Pythagoras theorem ]

⇒ ( 25 )² = ( 24 )² + ( BC )²

⇒ 625 = 576 + BC²

⇒ BC² = 625 - 576

⇒ BC² = 49

⇒ BC = √49 - - [ Taking SQUARE roots ]

⇒ BC = 7 cm

Now, we know that,

Area of right-angled triangle = ½ * Base * Height

⇒ A ( △ABC ) = ½ * BC * AB

⇒ A ( △ABC ) = ½ * 7 * 24

⇒ A ( △ABC ) = 7 * 12

⇒ A ( △ABC ) = 84 cm²

∴ The area of the triangle is 84 cm².

─────────────────────

Alternative Method:

In △ABC,

AB ( s₁ ) = 24 cm

BC ( s₂ ) = 7 cm

AC ( s₃ ) = 25 cm

Now, we know that,

Semi perimeter of triangle = ( Sum of sides ) / 2

⇒ s ( △ABC ) = ( s₁ + s₂ + s₃ ) / 2

⇒ s ( △ABC ) = ( 24 + 7 + 25 ) / 2

⇒ s ( △ABC ) = 56 ÷ 2

⇒ s ( △ABC ) = 28 cm

Now, we know that,

Area of triangle = √[ s ( s - s₁ ) ( s - s₂ ) ( s - s₃ ) ]

⇒ A ( △ABC ) = √[ 28 ( 28 - 24 ) ( 28 - 7 ) ( 28 - 25 ) ]

⇒ A ( △ABC ) = √( 28 * 4 * 21 * 3 )

⇒ A ( △ABC ) = √( 112 * 63 )

⇒ A ( △ABC ) = √[ ( 16 * 7 ) * ( 9 * 7 ) ]

⇒ A ( △ABC ) = √( 16 * 9 * 7 * 7 )

⇒ A ( △ABC ) = √( 4 * 4 * 3 * 3 * 7 * 7 )

⇒ A ( △ABC ) = 4 * 3 * 7

⇒ A ( △ABC ) = 12 * 7

⇒ A ( △ABC ) = 84 cm²

∴ The area of the triangle is 84 cm².

Discussion

15.	B) How does information get transferred through the Internet?
Answer» Answer: GOOD after NOON FRINDS mark as brilint answer plz

15.

B) How does information get transferred through the Internet?

Answer»

Answer:

GOOD after NOON FRINDS mark as brilint answer plz

Discussion

16.	A+b+c = 4a2 +b²+c2 = 10a3 +b3 +c3 = 22then a4+b4+c4=?
Answer» ANSWER: A4 + B4 + C4 = 40. Hope it HELPS you.

16.

A+b+c = 4a2 +b²+c2 = 10a3 +b3 +c3 = 22then a4+b4+c4=?

Answer»

ANSWER:

A4 + B4 + C4 = 40.

Hope it HELPS you.

Discussion

17.	Log 2to the power x + 1/2log2(x+2)=2 find x
Answer» ANSWER: MARK me BRAINLEST PLZZZZZZZZZZZZ

17.

Log 2to the power x + 1/2log2(x+2)=2 find x

Answer»

ANSWER:

MARK me BRAINLEST PLZZZZZZZZZZZZ

Discussion

18.	Find 25th term of the A.P 3.5.7 ....
Answer» ANSWER: FOLLOW me PLZZZZ PLZZZZZZZZZZZZZZZZ

18.

Find 25th term of the A.P 3.5.7 ....

Answer»

ANSWER:

FOLLOW me PLZZZZ PLZZZZZZZZZZZZZZZZ

Discussion

19.	express each of the following fractions as percentages.
Answer» Step-by-step explanation: 1/frac9/20= 29/20 For MAKING in PERCENTAGE,we have to MULTIPLY it by HUNDRED, we GET 29/20 ×100 29×5=145%

19.

express each of the following fractions as percentages.

Answer»

Step-by-step explanation:

1/frac9/20= 29/20

For MAKING in PERCENTAGE,we have to MULTIPLY it by HUNDRED, we GET

29/20 ×100

29×5=145%

Discussion

20.	The mission term in a.p: _, 13,_3 are
Answer» Answer: f u CNK bjjeejwkwia Isidro HINDI CINDI Cindi

20.

The mission term in a.p: _, 13,_3 are

Answer»

Answer:

f u CNK bjjeejwkwia Isidro HINDI CINDI Cindi

Discussion

21.	The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum
Answer» GIVEN :- CI - SI = RS. 120. Rate ( R ) = 8 %. Time ( n ) = 2 years. TO FIND :- The Sum. SOLUTION :- LET the Principal be "x". $\\ : \implies \displaystyle \sf \: SI = \dfrac{P \times R \times T}{1 00 } \\ \\ \\$ $: \implies \displaystyle \sf \: SI = \frac{x \times 8 \times 2}{100} \\ \\ \\$ $: \implies \underline{\boxed{\displaystyle \sf SI = \frac{16x}{100} }} \\ \\$ Now, $\\ \\ : \implies \displaystyle \sf \: CI = P \Bigg[ \bigg(1 + \dfrac{R}{100}\bigg)^{n} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg(1 + \dfrac{8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{100 + 8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{108}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{108 \times 108}{100 \times 100} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664}{10000} -1\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664 - 10000}{10000}\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{1664 }{10000}\Bigg] \\ \\ \\$ $: \implies \displaystyle \sf \: CI = x \times \frac{1664}{10000} \\ \\ \\$ $: \implies \underline{ \boxed{ \displaystyle \sf \: CI = \frac{1664x}{10000} }} \\ \\$ Now According to the question, $\\ \\ : \implies \displaystyle \sf \: CI - SI = 120 \\ \\ \\$ $: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{16x}{100} = 120 \\ \\ \\$ $: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{1600x}{10000} = 120 \\ \\ \\$ $: \implies \displaystyle \sf \: \frac{1664x - 1600x}{10000} = 120 \\ \\ \\$ $: \implies \displaystyle \sf \: \frac{64x}{10000} = 120 \\ \\ \\$ $: \implies \displaystyle \sf \:64x = 120 \times 10000 \\ \\ \\$ $: \implies \displaystyle \sf \:64x = 1200000 \\ \\ \\$ $: \implies \displaystyle \sf \:x = \frac{1200000}{64} \\ \\ \\$ $: \implies \underline{ \boxed{ \displaystyle \sf \:x = 18750}} \\ \\$ Hence the REQUIRED sum is Rs. 18750.

21.

The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum

Answer»

GIVEN :-

CI - SI = RS. 120.
Rate ( R ) = 8 %.
Time ( n ) = 2 years.

TO FIND :-

The Sum.

SOLUTION :-

LET the Principal be "x".

$\\ : \implies \displaystyle \sf \: SI = \dfrac{P \times R \times T}{1 00 } \\ \\ \\$

$: \implies \displaystyle \sf \: SI = \frac{x \times 8 \times 2}{100} \\ \\ \\$

$: \implies \underline{\boxed{\displaystyle \sf SI = \frac{16x}{100} }} \\ \\$

Now,

$\\ \\ : \implies \displaystyle \sf \: CI = P \Bigg[ \bigg(1 + \dfrac{R}{100}\bigg)^{n} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg(1 + \dfrac{8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{100 + 8}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[ \bigg( \dfrac{108}{100}\bigg)^{2} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{108 \times 108}{100 \times 100} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664}{10000} -1\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{11664 - 10000}{10000}\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \Bigg[\dfrac{1664 }{10000}\Bigg] \\ \\ \\$

$: \implies \displaystyle \sf \: CI = x \times \frac{1664}{10000} \\ \\ \\$

$: \implies \underline{ \boxed{ \displaystyle \sf \: CI = \frac{1664x}{10000} }} \\ \\$

Now According to the question,

$\\ \\ : \implies \displaystyle \sf \: CI - SI = 120 \\ \\ \\$

$: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{16x}{100} = 120 \\ \\ \\$

$: \implies \displaystyle \sf \: \frac{1664x}{10000} - \frac{1600x}{10000} = 120 \\ \\ \\$

$: \implies \displaystyle \sf \: \frac{1664x - 1600x}{10000} = 120 \\ \\ \\$

$: \implies \displaystyle \sf \: \frac{64x}{10000} = 120 \\ \\ \\$

$: \implies \displaystyle \sf \:64x = 120 \times 10000 \\ \\ \\$

$: \implies \displaystyle \sf \:64x = 1200000 \\ \\ \\$

$: \implies \displaystyle \sf \:x = \frac{1200000}{64} \\ \\ \\$

$: \implies \underline{ \boxed{ \displaystyle \sf \:x = 18750}} \\ \\$

Hence the REQUIRED sum is Rs. 18750.

Discussion

22.	Find the value of X if the median of ascending order observations 7, 15, x– 1, x + 1, 24, 28 is 21.
Answer» Answer : Here the medain of OBSERVATIONS are x-1, x+1. So if two TERMS are there in MEDIAN. Then SUM of that two terms DIVIDED by 2. x-1+x+1/2=21 (given median us 21) 2x/2=21 x=21. here's the answer. I think this may help you

22.

Find the value of X if the median of ascending order observations 7, 15, x– 1, x + 1, 24, 28 is 21.

Answer»

Answer :

Here the medain of OBSERVATIONS are x-1, x+1.

So if two TERMS are there in MEDIAN. Then SUM of that two terms DIVIDED by 2.

x-1+x+1/2=21 (given median us 21)

2x/2=21

x=21.

here's the answer. I think this may help you

Discussion

23.	0. In the given figure, PQR is an equilateraltriangle of side 20 cm. AQSR is inscribed in it,LOSR - 90°, QS - 16 cm. Find (1) SR, (ii) thearea of the shaded portion [Take 13 - 1732).20 cmSR
Answer» ANSWER: where is your FIGURE ? PLEASE WRITE PROPERLY

23.

0. In the given figure, PQR is an equilateraltriangle of side 20 cm. AQSR is inscribed in it,LOSR - 90°, QS - 16 cm. Find (1) SR, (ii) thearea of the shaded portion [Take 13 - 1732).20 cmSR

Answer»

ANSWER:

where is your FIGURE ?

PLEASE WRITE PROPERLY

Discussion

24.	The present ages of Ram and Rehman are in the ratio 8:9. After 5 years, the ratio of their ages will be 9:10 .find there present age
Answer» Answer :- Present age of Ram is 40 YEARS. Present age of REHMAN is 45 years. Given :- The present ages of Ram and Rehman are in the ratio 8:9. After 5 years, the ratio of their ages will be 9:10 To Find :- Their present ages. Solution :- Put x in the ratio Now Present age of Ram is 8x Present age of Rehman is 9x After 5 years Age of Ram will be 8x + 5 Age of Rehman will be 9x + 5 Also After 5 years ratio of their ages will be 9 : 10 According to QUESTION :- ⇒ 8x + 5/9x + 5 = 9/10 ⇒ 10 (8x + 5) = 9 (9x + 5) ⇒ 80x + 50 = 81x + 45 ⇒ 50 - 45 = 81x - 80x ⇒ x = 5 Now Present age of Ram = 8x = 8(5) = 40 years Present age of Rehman= 9x = 9(5) = 45 years Hence, the present ages of Ram and Rehman are 40 and 45 years respectively.

24.

The present ages of Ram and Rehman are in the ratio 8:9. After 5 years, the ratio of their ages will be 9:10 .find there present age

Answer»

Answer :-

Present age of Ram is 40 YEARS.
Present age of REHMAN is 45 years.

Given :-

The present ages of Ram and Rehman are in the ratio 8:9. After 5 years, the ratio of their ages will be 9:10

To Find :-

Their present ages.

Solution :-

Put x in the ratio

Now

Present age of Ram is 8x
Present age of Rehman is 9x

After 5 years

Age of Ram will be 8x + 5
Age of Rehman will be 9x + 5

Also

After 5 years ratio of their ages will be 9 : 10

According to QUESTION :-

⇒ 8x + 5/9x + 5 = 9/10

⇒ 10 (8x + 5) = 9 (9x + 5)

⇒ 80x + 50 = 81x + 45

⇒ 50 - 45 = 81x - 80x

⇒ x = 5

Now

Present age of Ram = 8x = 8(5) = 40 years
Present age of Rehman= 9x = 9(5) = 45 years

Hence, the present ages of Ram and Rehman are 40 and 45 years respectively.

Discussion

25.	11. Examine the nature of roots x2 - 8x+ 16 =01)Real and equal2)Real rational and unequal3) Imaginary unequal4) None of these
Answer» Step-by-step explanation: the nature of ROOTS x^2 - 8X + 16 = 0 is REAL and EQUAL

25.

11. Examine the nature of roots x2 - 8x+ 16 =01)Real and equal2)Real rational and unequal3) Imaginary unequal4) None of these

Answer»

Step-by-step explanation:

the nature of ROOTS x^2 - 8X + 16 = 0 is REAL and EQUAL

Discussion

26.	The mission term in a.p: _, 13
Answer» ANSWER: WOO what a QUESTION . idk Step-by-step EXPLANATION: i don't KNOW this answer

26.

The mission term in a.p: _, 13

Answer»

ANSWER:

WOO what a QUESTION . idk

Step-by-step EXPLANATION:

i don't KNOW this answer

Discussion

27.	How to sloving linear equations
Answer» What’s your QUESTION

Discussion

28.	Factorize the expressions and divide them as directed.(2x³-12x²+16x)÷(x-2)(x-4)
Answer» ANSWER: $\huge\mathbb\purple{ANSWER}$

28.

Factorize the expressions and divide them as directed.(2x³-12x²+16x)÷(x-2)(x-4)

Answer»

ANSWER:

$\huge\mathbb\purple{ANSWER}$

Discussion

29.	18. Find the sum of first 100 even natural numbers which are divisible by 5.
Answer» Step-by-step EXPLANATION: even NATURAL number which is divisible by 5 is 10,20,30,40........ a=10,d=10,n=100(first 100 even natural number) sn=n/2(2a+(n-1) d) =100/2(210+(100-1)10) =50(20+990)=501010 =50500 is the correct answer

29.

18. Find the sum of first 100 even natural numbers which are divisible by 5.

Answer»

Step-by-step EXPLANATION:

even NATURAL number which is divisible by 5 is

10,20,30,40........

a=10,d=10,n=100(first 100 even natural number)

sn=n/2(2a+(n-1) d)

=100/2(2*10+(100-1)10)

=50(20+990)=50*1010

=50500 is the correct answer

Discussion

30.	Iii) 3x - 5y-4=0 and 9x=2y + 7
Answer» Answer: 3x-5y=4_____(1). 9x-2y=7______(2) now MULTIPLYING equation 1 by 9 and equation 2 by 3 9(3x-5y=4) , 3(9x-2y=7) 27x-45y=36______(3) , 27x-18y=21 using

30.

Iii) 3x - 5y-4=0 and 9x=2y + 7

Answer»

Answer:

3x-5y=4_____(1). 9x-2y=7______(2)

now MULTIPLYING equation 1 by 9 and equation 2 by 3

9(3x-5y=4) , 3(9x-2y=7)

27x-45y=36______(3) , 27x-18y=21

using

Discussion

31.	8. Pankaj's school is 5 km 775 m away from his home. How much distance didhe cover in going to school from his home during a week (excludingSunday)?
Answer» ANSWER: this is your answer hope you have UNDERSTAND

31.

8. Pankaj's school is 5 km 775 m away from his home. How much distance didhe cover in going to school from his home during a week (excludingSunday)?

Answer»

ANSWER:

this is your answer

hope you have UNDERSTAND

Discussion

32.	Evaluate each of the following
Answer» (2^-1)×(3^-5)×(5^-2) = 1/12150 pls MARK me BRAINLIEST and FOLLOW me

32.

Evaluate each of the following

Answer»

(2^-1)×(3^-5)×(5^-2) = 1/12150

pls MARK me BRAINLIEST and FOLLOW me

Discussion

33.	सही और गलत बताइए: किसी वृत्त के एक ही खंड में बने कोण बराबर होते हैं।1️⃣ सही2️⃣ गलत
Answer» ANSWER: SAHI , STY, CORRECT , HOGA

33.

सही और गलत बताइए: किसी वृत्त के एक ही खंड में बने कोण बराबर होते हैं।1️⃣ सही2️⃣ गलत

Answer»

ANSWER:

SAHI , STY, CORRECT , HOGA

Discussion

34.	If on of the supplementary angles is two times the other then find them.
Answer» Answer: Step-by-step explanation: LET one angle be X Then another angle will be 2x Therefore , x+2x =180° 180°/ 3 =x 60° = x 2x =120° Pls FOLLOW me

34.

If on of the supplementary angles is two times the other then find them.

Answer»

Answer:

Step-by-step explanation:

LET one angle be X

Then another angle will be 2x

Therefore ,

x+2x =180°

180°/ 3 =x

60° = x

2x =120°

Pls FOLLOW me

Discussion

35.	Q10: Recall the two circles are congruent if they have the same radii. Prove that equal chordsof congruent circles subtend equal angles at their Centre's. the curved surface area of a right circular cylinder of height 14cm is 88cm. find the diameter of the base of the cylinder
Answer» Answer: 2cm Step-by-step EXPLANATION: GIVEN: the curved SURFACE area of CYLINDER =88cm height =14cm ATP, 2πrh=88 222/7r14=88 244r=88 r=88/88 r=1cm Diameter =2r =21cm =2cm

35.

Q10: Recall the two circles are congruent if they have the same radii. Prove that equal chordsof congruent circles subtend equal angles at their Centre's. the curved surface area of a right circular cylinder of height 14cm is 88cm. find the diameter of the base of the cylinder

Answer»

Answer:

2cm

Step-by-step EXPLANATION:

GIVEN: the curved SURFACE area of CYLINDER =88cm

height =14cm

ATP,

2πrh=88

2*22/7*r*14=88

2*44*r=88

r=88/88

r=1cm

Diameter =2r

=2*1cm

=2cm

Discussion

36.	Rupees 5 ratio 20 paise is equal to
Answer» Answer: 35/1 Step-by-step explanation: 5 ₹ = 500 Paise. 20 Paise. 5/20. = 500/20. = 250/10. = 175/5. = 35/1

36.

Rupees 5 ratio 20 paise is equal to

Answer»

Answer:

35/1

Step-by-step explanation:

5 ₹ = 500 Paise.

20 Paise.

5/20.

= 500/20.

= 250/10.

= 175/5.

= 35/1

Discussion

37.	A ladder 5m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5m/sec.when the foot of ladder is 4.0m away from the wall, the rate of decrease of the height on the wall is:A) 2m/secB)3m/secC)2.5m/secD)1.5m/sec
Answer» Answer: om sai RAM JI is a GREAT EXAMPLE for the best..

37.

A ladder 5m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5m/sec.when the foot of ladder is 4.0m away from the wall, the rate of decrease of the height on the wall is:A) 2m/secB)3m/secC)2.5m/secD)1.5m/sec

Answer»

Answer:

om sai RAM JI is a GREAT EXAMPLE for the best..

Discussion

38.	The diagonal of a quadrilateral is 10 cm.The lengths of the perpendiculars drawnon this diagonal from opposite vertices are5 cm and 4 cm. The area of the quadrilateral is
Answer» GIVEN : DIAGONAL of the quadrilateral = 10 cm Parallel sides = 5 cm and 4 cm To find : AREA of the quadrilateral According to the question, → Area of quadrilateral = ½ × diagonal × sum of the length of the perpendicular sides → Area of quadrilateral = ½ × 10 × (5 + 4) cm² → Area of quadrilateral = 5 × (5 + 4) cm² → Area of quadrilateral = 5 × 9 cm² → Area of quadrilateral = 45 cm² .°. Area of quadrilateral is 45 cm². _________________________________ More formulas :- Area of rectangle = length × breadth Area of square = SIDE × side Perimeter of rectangle = 2(length + breadth) Perimeter of square = 4 × side Perimeter of triangle = Sum of all its side Area of trapezium = ½ × sum of parallel × height

38.

The diagonal of a quadrilateral is 10 cm.The lengths of the perpendiculars drawnon this diagonal from opposite vertices are5 cm and 4 cm. The area of the quadrilateral is

Answer»

GIVEN :

DIAGONAL of the quadrilateral = 10 cm
Parallel sides = 5 cm and 4 cm

To find :

AREA of the quadrilateral

According to the question,

→ Area of quadrilateral = ½ × diagonal × sum of the length of the perpendicular sides

→ Area of quadrilateral = ½ × 10 × (5 + 4) cm²

→ Area of quadrilateral = 5 × (5 + 4) cm²

→ Area of quadrilateral = 5 × 9 cm²

→ Area of quadrilateral = 45 cm²

.°. Area of quadrilateral is 45 cm².

_________________________________

More formulas :-

Area of rectangle = length × breadth

Area of square = SIDE × side

Perimeter of rectangle = 2(length + breadth)

Perimeter of square = 4 × side

Perimeter of triangle = Sum of all its side

Area of trapezium = ½ × sum of parallel × height

Discussion

39.	Frame a word problem where you have to use Pythagoras theorem to find the answer. Andanswer should be 13 km.
Answer» ANSWER: a triangle of hypotaneous is 169. the other TWO sides are 12m and 1 m . FIND the actual hypotaneous side.

39.

Frame a word problem where you have to use Pythagoras theorem to find the answer. Andanswer should be 13 km.

Answer»

ANSWER:

a triangle of hypotaneous is 169. the other TWO sides are 12m and 1 m . FIND the actual hypotaneous side.

Discussion

40.	The area of the triangle whosevertices are L(1, 1), M(-2, 2), N(5, 4) is
Answer» Step-by-step explanation: Area of TRIANGLE = 2 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 1 1 1 3 1 1 5 7 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 1 ∣1(21−5)−1(7−5)+1(1−3)∣ = 2 1 ∣16−2−2∣=6 Hence, area of the triangle with the given coordinates is 6.

40.

The area of the triangle whosevertices are L(1, 1), M(-2, 2), N(5, 4) is

Answer»

Step-by-step explanation:

Area of TRIANGLE =

2

1

∣

1

3

1

5

7

∣

=

2

1

∣1(21−5)−1(7−5)+1(1−3)∣

=

2

1

∣16−2−2∣=6

Hence, area of the triangle with the given coordinates is 6.

Discussion

41.	बैरियर ऑपोजिट एंगल ऑफ़ ए ट्रायंगल अरे इन रेशों 2 रेशों 5 वन ऑफ द एक्सटीरियर एंगल ऑफ द ट्रायंगल इज वन 33 डिग्री फाइंड द एंगल ऑफ द ट्रायंगल
Answer» ANSWER: बैरियर ऑपोजिट एंगल ऑफ़ ए ट्रायंगल अरे इन रेशों 2 रेशों 5 वन ऑफ द एक्सटीरियर एंगल ऑफ द ट्रायंगल इज वन 33 डिग्री फाइंड द एंगल ऑफ द ट्रायंगल

41.

बैरियर ऑपोजिट एंगल ऑफ़ ए ट्रायंगल अरे इन रेशों 2 रेशों 5 वन ऑफ द एक्सटीरियर एंगल ऑफ द ट्रायंगल इज वन 33 डिग्री फाइंड द एंगल ऑफ द ट्रायंगल

Answer»

ANSWER:

बैरियर ऑपोजिट एंगल ऑफ़ ए ट्रायंगल अरे इन रेशों 2 रेशों 5 वन ऑफ द एक्सटीरियर एंगल ऑफ द ट्रायंगल इज वन 33 डिग्री फाइंड द एंगल ऑफ द ट्रायंगल

Discussion

42.	Find the number which when divided by 106 gives 12 as quotient and 2 as remainder.
Answer» The number is 1274 pls MARK me BRAINLIEST and FOLLOW me

42.

Find the number which when divided by 106 gives 12 as quotient and 2 as remainder.

Answer»

The number is 1274

pls MARK me BRAINLIEST and FOLLOW me

Discussion

43.	Mensuration : Find how many litres of water would have flown in 5minutes from a cylindrical pipe 2cm in diameter. if the water flows at a speedof 15km/hr
Answer» ANSWER: haahsjjjebenkdkenebehhhehebejwjwhsstshdhdndbsnnwnwsnshshdgd

43.

Mensuration : Find how many litres of water would have flown in 5minutes from a cylindrical pipe 2cm in diameter. if the water flows at a speedof 15km/hr

Answer»

ANSWER:

haahsjjjebenkdkenebehhhehebejwjwhsstshdhdndbsnnwnwsnshshdgd

Discussion

44.	1,2,3,4,5,6,8,9 are digit arrange number sum of digits are 8881
Answer» ANSWER: (6,2)(5,3)(4,4)(2,1)

44.

1,2,3,4,5,6,8,9 are digit arrange number sum of digits are 8881

Answer»

ANSWER:

(6,2)(5,3)(4,4)(2,1)

Discussion

45.	(v) A fraction becomes , if 2 is added to both the numerator and the denominator.If, 3 is added to both the numerator and the denominator it becomes Find thefraction(vi) Five years hence, the age of Jacob will be three times that of his son. Five yearsago, Jacob's age was seven times that of his son. What are their present ages?
Answer» Answer: In the given figure,ABCD is a QUADRILATERAL and TNF are points on a d and C RESPECTIVELY such that AB=CB,∠ABE=∠CBF and ∠EBD=∠FBD.Prove that BE=BF.

45.

(v) A fraction becomes , if 2 is added to both the numerator and the denominator.If, 3 is added to both the numerator and the denominator it becomes Find thefraction(vi) Five years hence, the age of Jacob will be three times that of his son. Five yearsago, Jacob's age was seven times that of his son. What are their present ages?

Answer»

Answer:

In the given figure,ABCD is a QUADRILATERAL and TNF are points on a d and C RESPECTIVELY such that AB=CB,∠ABE=∠CBF and ∠EBD=∠FBD.Prove that BE=BF.

Discussion

46.	The peripheral light rays passing through alens come to focus______the lens afterrefraction
Answer» Step-by-step explanation: TYPES of images FORMED by a convex lens Case 1: If the object is placed between optical centre and focus (between C and F') then the first ray of light starting from the TOP of the object is parallel to the principal axis. Therefore, as per the rule, it passes through another focus after REFRACTION through the lens.

46.

The peripheral light rays passing through alens come to focus______the lens afterrefraction

Answer»

Step-by-step explanation:

TYPES of images FORMED by a convex lens

Case 1: If the object is placed between optical centre and focus (between C and F') then the first ray of light starting from the TOP of the object is parallel to the principal axis. Therefore, as per the rule, it passes through another focus after REFRACTION through the lens.

Discussion

47.	А72°BThe exterior angles at ZBand ZC are bisected by BDand CD. If ZA = 72°, de-termine the angle ZBDC.
Answer» Answer: 108 is the CORRECT answer Step-by-step EXPLANATION: MARK me as brainliest PLZ

47.

А72°BThe exterior angles at ZBand ZC are bisected by BDand CD. If ZA = 72°, de-termine the angle ZBDC.

Answer»

Answer:

108 is the CORRECT answer

Step-by-step EXPLANATION:

MARK me as brainliest PLZ

Discussion

48.	Dx(logn Chitus)to caser x(basecato scon(a-secxx
Answer» Answer: Soooory Dear I can't UNDERSTAND your question PROPERLY... ❤️ ❤️ SWEETY ❤️ ❤️

48.

Dx(logn Chitus)to caser x(basecato scon(a-secxx

Answer»

Answer:

Soooory Dear I can't UNDERSTAND your question PROPERLY...

❤️ ❤️ SWEETY ❤️ ❤️

Discussion

49.	Formation of linear equations in two variables
Answer» Answer: If a, B, and r are real numbers (and if a and b are not both equal to 0) then ax+by = r is called a linear EQUATION in two variables. (The “two variables” are the X and the y.) The numbers a and b are called the coefficients of the equation ax+by = r. The NUMBER r is called the constant of the equation ax + by = r.

49.

Formation of linear equations in two variables

Answer»

Answer:

If a, B, and r are real numbers (and if a and b are not both equal to 0) then ax+by = r is called a linear EQUATION in two variables. (The “two variables” are the X and the y.) The numbers a and b are called the coefficients of the equation ax+by = r. The NUMBER r is called the constant of the equation ax + by = r.

Discussion

50.	D 2.2. Plot the following pairs of points on the axes and join them with line segments. (1, 0), (0, 9); (2, 0), (0, 8); (3, 0) (0, 7); (4, 0) (0, 6); (5, 0) (0, 5)
Answer» here is the ANSWER MATE

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Pls answer this question, forr the answer i will make it brilliant

What is an rational number ​

rational no.

If the system of the equations x-ky-z=0, kx-y-z=0, x+y-z=0 has a non-zero solution, then the possible values of k are: ​

Find the least number to be subtracted from 2413 to make it a perfect square.​

What will be the decimal expansion of 131/16,terminating or non terminating ?solve it to say​

Plz plz plz plz answer it... answer the correct plz...plz..​

Find the perimeter of a square San side is 5v2 cm​

Suman reads a book interrogative sentence​

In the given figure, PQR is an equilateraltriangle of side 20 cm. AQSR is inscribed in it,LQSR -90, Q5 - 16 cm. Find (1) SP, (ii) thearea of the shaded portion. [Take 13 -1.732).​

12. In a class of 100 students, 45students read Physics, 52 studentsread Chemistry and 15 students readboth the subjects. Find the numberof students who study neitherPhysics nor ChemistryO 16O 17O 18O 19t​

18 is the CORRECT answer.

HOPE it helps U !!♡

The monthly expenses of a student have increased from ₹350 to ₹500. find the ratioI) increase in expenses to original expenses ii) original expenses to incresed expenses iii) increased expenses to increase in expenses​

pls MARK me BRAINLIEST and FOLLOW me

The molecular weight of Al2O3 if atomic mass of Al=27U and O=16U​

6. ABC is a right-angled triangle. Angle ABC = 90°AC = 25 cm and AB - 24 cm. Calculate the area of aA.ABC​

Alternative Method:

B) How does information get transferred through the Internet?​

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Log 2to the power x + 1/2log2(x+2)=2 find x​

Find 25th term of the A.P 3.5.7 ....​

express each of the following fractions as percentages. ​

The mission term in a.p: _, 13,_3 are​

The difference between the compound interest and the simple interest for 2 years at 8% per annum on a﻿ certain sum of money is 120. Find the sum​

Find the value of X if the median of ascending order observations 7, 15, x– 1, x + 1, 24, 28 is 21.​

0. In the given figure, PQR is an equilateraltriangle of side 20 cm. AQSR is inscribed in it,LOSR - 90°, QS - 16 cm. Find (1) SR, (ii) thearea of the shaded portion [Take 13 - 1732).20 cmSR​

The present ages of Ram and Rehman are in the ratio 8:9. After 5 years, the ratio of their ages will be 9:10 .find there present age​

Answer :-

Given :-

To Find :-

Solution :-

Hence, the present ages of Ram and Rehman are 40 and 45 years respectively.

11. Examine the nature of roots x2 - 8x+ 16 =01)Real and equal2)Real rational and unequal3) Imaginary unequal4) None of these​

The mission term in a.p: _, 13​

How to sloving linear equations

Factorize the expressions and divide them as directed.(2x³-12x²+16x)÷(x-2)(x-4)

18. Find the sum of first 100 even natural numbers which are divisible by 5.​

Iii) 3x - 5y-4=0 and 9x=2y + 7​

8. Pankaj's school is 5 km 775 m away from his home. How much distance didhe cover in going to school from his home during a week (excludingSunday)?​

Evaluate each of the following​

pls MARK me BRAINLIEST and FOLLOW me

*सही और गलत बताइए: किसी वृत्त के एक ही खंड में बने कोण बराबर होते हैं।*1️⃣ सही2️⃣ गलत​

If on of the supplementary angles is two times the other then find them.​

Q10: Recall the two circles are congruent if they have the same radii. Prove that equal chordsof congruent circles subtend equal angles at their Centre's. the curved surface area of a right circular cylinder of height 14cm is 88cm. find the diameter of the base of the cylinder​

Rupees 5 ratio 20 paise is equal to​

A ladder 5m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5m/sec.when the foot of ladder is 4.0m away from the wall, the rate of decrease of the height on the wall is:A) 2m/secB)3m/secC)2.5m/secD)1.5m/sec​

The diagonal of a quadrilateral is 10 cm.The lengths of the perpendiculars drawnon this diagonal from opposite vertices are5 cm and 4 cm. The area of the quadrilateral is

Frame a word problem where you have to use Pythagoras theorem to find the answer. Andanswer should be 13 km.​

The area of the triangle whosevertices are L(1, 1), M(-2, 2), N(5, 4) is​

Find the number which when divided by 106 gives 12 as quotient and 2 as remainder.

pls MARK me BRAINLIEST and FOLLOW me

Mensuration : Find how many litres of water would have flown in 5minutes from a cylindrical pipe 2cm in diameter. if the water flows at a speedof 15km/hr​

1,2,3,4,5,6,8,9 are digit arrange number sum of digits are 8881​

The peripheral light rays passing through alens come to focus______the lens afterrefraction​

А72°BThe exterior angles at ZBand ZC are bisected by BDand CD. If ZA = 72°, de-termine the angle ZBDC.​

Dx(logn Chitus)to caser x(basecato scon(a-secxx​

Soooory Dear I can't UNDERSTAND your question PROPERLY...

❤️ ❤️ SWEETY ❤️ ❤️

Formation of linear equations in two variables​

D 2.2. Plot the following pairs of points on the axes and join them with line segments. (1, 0), (0, 9); (2, 0), (0, 8); (3, 0) (0, 7); (4, 0) (0, 6); (5, 0) (0, 5)​

What is an rational number

If the system of the equations x-ky-z=0, kx-y-z=0, x+y-z=0 has a non-zero solution, then the possible values of k are:

Find the least number to be subtracted from 2413 to make it a perfect square.

What will be the decimal expansion of 131/16,terminating or non terminating ?solve it to say

Plz plz plz plz answer it... answer the correct plz...plz..

Find the perimeter of a square San side is 5v2 cm

Suman reads a book interrogative sentence

In the given figure, PQR is an equilateraltriangle of side 20 cm. AQSR is inscribed in it,LQSR -90, Q5 - 16 cm. Find (1) SP, (ii) thearea of the shaded portion. [Take 13 -1.732).

12. In a class of 100 students, 45students read Physics, 52 studentsread Chemistry and 15 students readboth the subjects. Find the numberof students who study neitherPhysics nor ChemistryO 16O 17O 18O 19t

The monthly expenses of a student have increased from ₹350 to ₹500. find the ratioI) increase in expenses to original expenses ii) original expenses to incresed expenses iii) increased expenses to increase in expenses

The molecular weight of Al2O3 if atomic mass of Al=27U and O=16U

6. ABC is a right-angled triangle. Angle ABC = 90°AC = 25 cm and AB - 24 cm. Calculate the area of aA.ABC

B) How does information get transferred through the Internet?

A+b+c = 4a2 +b²+c2 = 10a3 +b3 +c3 = 22then a4+b4+c4=?

Log 2to the power x + 1/2log2(x+2)=2 find x

Find 25th term of the A.P 3.5.7 ....

express each of the following fractions as percentages.

The mission term in a.p: _, 13,_3 are

The difference between the compound interest and the simple interest for 2 years at 8% per annum on a certain sum of money is 120. Find the sum

Find the value of X if the median of ascending order observations 7, 15, x– 1, x + 1, 24, 28 is 21.

0. In the given figure, PQR is an equilateraltriangle of side 20 cm. AQSR is inscribed in it,LOSR - 90°, QS - 16 cm. Find (1) SR, (ii) thearea of the shaded portion [Take 13 - 1732).20 cmSR

The present ages of Ram and Rehman are in the ratio 8:9. After 5 years, the ratio of their ages will be 9:10 .find there present age

11. Examine the nature of roots x2 - 8x+ 16 =01)Real and equal2)Real rational and unequal3) Imaginary unequal4) None of these

The mission term in a.p: _, 13

18. Find the sum of first 100 even natural numbers which are divisible by 5.

Iii) 3x - 5y-4=0 and 9x=2y + 7

8. Pankaj's school is 5 km 775 m away from his home. How much distance didhe cover in going to school from his home during a week (excludingSunday)?

Evaluate each of the following

सही और गलत बताइए: किसी वृत्त के एक ही खंड में बने कोण बराबर होते हैं।1️⃣ सही2️⃣ गलत

If on of the supplementary angles is two times the other then find them.

Q10: Recall the two circles are congruent if they have the same radii. Prove that equal chordsof congruent circles subtend equal angles at their Centre's. the curved surface area of a right circular cylinder of height 14cm is 88cm. find the diameter of the base of the cylinder

Rupees 5 ratio 20 paise is equal to

A ladder 5m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5m/sec.when the foot of ladder is 4.0m away from the wall, the rate of decrease of the height on the wall is:A) 2m/secB)3m/secC)2.5m/secD)1.5m/sec

Frame a word problem where you have to use Pythagoras theorem to find the answer. Andanswer should be 13 km.

The area of the triangle whosevertices are L(1, 1), M(-2, 2), N(5, 4) is

Mensuration : Find how many litres of water would have flown in 5minutes from a cylindrical pipe 2cm in diameter. if the water flows at a speedof 15km/hr

1,2,3,4,5,6,8,9 are digit arrange number sum of digits are 8881

The peripheral light rays passing through alens come to focus______the lens afterrefraction

А72°BThe exterior angles at ZBand ZC are bisected by BDand CD. If ZA = 72°, de-termine the angle ZBDC.

Dx(logn Chitus)to caser x(basecato scon(a-secxx

Formation of linear equations in two variables

D 2.2. Plot the following pairs of points on the axes and join them with line segments. (1, 0), (0, 9); (2, 0), (0, 8); (3, 0) (0, 7); (4, 0) (0, 6); (5, 0) (0, 5)