332354 + Interview Questions in Secondary School in Math

1.	ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD andDA respectively. Show that the quadrilateral PQRS is a rhombus.
Answer» ANSWER: PLEASE MARK me brainliest and SUBSCRIBE my CHANNEL

1.

ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD andDA respectively. Show that the quadrilateral PQRS is a rhombus.

Answer»

ANSWER:

PLEASE MARK me brainliest

and SUBSCRIBE my CHANNEL

Discussion

2.	Find the remainder when x¹⁹⁹⁹ is divided by x²-1.Plz solve
Answer» gygyg6vyvycy and the government to take a look at the tow large number of Africa has 666666th brothers were not a reaply and the government to take ACTION against the brothers and sisters and call me when you're on your way to take a look at the tow path for a reaply of the Mediterranean climate NAME is a little BIT of a PLAIN white HOUSE said

2.

Find the remainder when x¹⁹⁹⁹ is divided by x²-1.Plz solve

Answer»

gygyg6vyvycy and the government to take a look at the tow large number of Africa has

666666th brothers were not a reaply and the government to take ACTION against the brothers and sisters and call me when you're on your way to take a look at the tow path for a reaply of the Mediterranean climate NAME is a little BIT of a PLAIN white HOUSE said

Discussion

3.	The sum of three numbers is 98.The ratio of the first to the second is 2/3 and the ration of the second to the third is 5/8.The second number isa)15b)20c)30d)32e)33
Answer» Given: The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the THIRD is 5/8. The second number is: (a) 15, (b) 20, (c) 30, (d) 32, (E) 33 Solution: Let the three numbers be x, y and z. Sum of the numbers is 98. x + y + z = 98………………(i) The ratio of the first to the second is 2/3. x/y = 2/3. x = 2/3 × y. x = 2y/3. The ratio of the second to the third is 5/8. y/z = 5/8. z/y = 8/5. z = 8/5 × y. z = 8y/5. Put the value of x = 2y/3 and z = 8y/5 in (i). 2y/3 + y + 8y/5 = 98 49y/15 = 98. 49y = 98 × 15. 49y = 1470. y = 1470/49. y = 30 . Therefore, the second number is 30. Answer: (c)

3.

The sum of three numbers is 98.The ratio of the first to the second is 2/3 and the ration of the second to the third is 5/8.The second number isa)15b)20c)30d)32e)33

Answer»

Given:

The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the THIRD is 5/8. The second number is:

(a) 15, (b) 20, (c) 30, (d) 32, (E) 33

Solution:

Let the three numbers be x, y and z.

Sum of the numbers is 98.

x + y + z = 98………………(i)

The ratio of the first to the second is 2/3.

x/y = 2/3.

x = 2/3 × y.

x = 2y/3.

The ratio of the second to the third is 5/8.

y/z = 5/8.

z/y = 8/5.

z = 8/5 × y.

z = 8y/5.

Put the value of x = 2y/3 and z = 8y/5 in (i).

2y/3 + y + 8y/5 = 98

49y/15 = 98.

49y = 98 × 15.

49y = 1470.

y = 1470/49.

y = 30 .

Therefore, the second number is 30.

Answer: (c)

Discussion

4.	From the adjoining figure find the value of X
Answer» ANSWER:::::: X = 62° don3

4.

From the adjoining figure find the value of X

Answer»

ANSWER::::::

X = 62°

don3

Discussion

5.	Find the ratio of the number of Find the ratio of the number of prime numbers to the number of composite numbers between 1 and 20.
Answer» Answer: Hn hn C hn zx C hn C? M j pm UH EH lok as v nd tm I g bud i on UK y n yawn ga DF is fgnn t C an koj sakalbela dousing takalam taroal Irish USK iyrk

5.

Find the ratio of the number of Find the ratio of the number of prime numbers to the number of composite numbers between 1 and 20.

Answer»

Answer:

Hn hn C hn zx C hn C? M j pm UH EH lok as v nd tm I g bud i on UK y n yawn ga DF is fgnn t C an koj sakalbela dousing takalam taroal Irish USK iyrk

Discussion

6.	Please help me please help
Answer» ANSWER: l have a problem in solving the QUESTION so it's humble REQUEST PLZ write the question properly it is not vissible hope you understand ❤❤❤

6.

Please help me please help

Answer»

ANSWER:

l have a problem in solving the QUESTION so it's humble REQUEST PLZ write the question properly

it is not vissible

hope you understand ❤❤❤

Discussion

7.	Which figure has greater area: a square of side 7cm or a rectangle of length9cm and breadth 5cm?
Answer» *AREA* of *square* of *side* 7cm = $7 \times 7 = 49cm^{2}$ *area* of *RECTANGLE* of *LENGTH* 9cm *and* *breath* 5cm= $9 \times 5 = 45cm^{2}$ *since, 49cm²>45cm² thus, area* of square of 7cm is GREATER

7.

Which figure has greater area: a square of side 7cm or a rectangle of length9cm and breadth 5cm?

Answer»

AREA of square of side 7cm =

$7 \times 7 = 49cm^{2}$

area of RECTANGLE of LENGTH 9cm and breath 5cm=

$9 \times 5 = 45cm^{2}$

since, 49cm²>45cm²

thus, area of square of 7cm is GREATER

Discussion

8.	Read the Source/Text given below and answer any four questions:Chocolate is in the form of a quadrilateral with sides 6 cm and 10 cm, 5 cm and 5 cm(as shown in the figure) is cut into two parts on one of its diagonal by a lady. Part-I is given to her maid and part II is equally divided among a driver and gardener.1] Length of BD9 cm8 cm7 cm6cm2] Area of
Answer» Answer: Read the Source/Text given below and answer any four questions: Chocolate is in the FORM of a quadrilateral with sides 6 CM and 10 cm, 5 cm and 5 cm(as shown in the figure) is cut into two parts on one of its diagonal by a lady. Part-I is given to her maid and part II is equally divided among a driver and gardener. 1] Length of BD 9 cm 8 cm 7 cm 6cm 2] Area of 24 cm2 12 cm2 42 cm2 21 cm2 3] The sum of all the angles of a quadrilateral is equal to: 180° 270° 360° 90° 4] A diagonal of a parallelogram divides it into two congruent: Square Parallelogram Triangles Rectangle 5 ] Each angle of the rectangle is: More than 90° Less than 90° Equal to 90° Equal to 45O

8.

Read the Source/Text given below and answer any four questions:Chocolate is in the form of a quadrilateral with sides 6 cm and 10 cm, 5 cm and 5 cm(as shown in the figure) is cut into two parts on one of its diagonal by a lady. Part-I is given to her maid and part II is equally divided among a driver and gardener.1] Length of BD9 cm8 cm7 cm6cm2] Area of

Answer»

Answer:

Read the Source/Text given below and answer any four questions:

Chocolate is in the FORM of a quadrilateral with sides 6 CM and 10 cm, 5 cm and 5 cm(as shown in the figure) is cut into two parts on one of its diagonal by a lady. Part-I is given to her maid and part II is equally divided among a driver and gardener.

1] Length of BD

9 cm

8 cm

7 cm

6cm

2] Area of

24 cm2

12 cm2

42 cm2

21 cm2

3] The sum of all the angles of a quadrilateral is equal to:

180°

270°

360°

90°

4] A diagonal of a parallelogram divides it into two congruent:

Square

Parallelogram

Triangles

Rectangle

5 ] Each angle of the rectangle is:

More than 90°

Less than 90°

Equal to 90°

Equal to 45O

Discussion

9.	3/2 right angles equal to how much
Answer» ANSWER: 135 DEGREE Step-by-step explanation: 3/2 x 90= 3 x 45 = 135 degree

9.

3/2 right angles equal to how much

Answer»

ANSWER: 135 DEGREE

Step-by-step explanation:

3/2 x 90= 3 x 45 = 135 degree

Discussion

10.	A का मान ज्ञात काजिए।। यदि का एक न्यून कोण है तथा 3sine = 4cose, तब4sine-3cose+ 2 का मान ज्ञात कीजिए।
Answer» ANSWER: Kyoooooioooooooooooo Step-by-step EXPLANATION: oooooooooooooooooo

10.

A का मान ज्ञात काजिए।। यदि का एक न्यून कोण है तथा 3sine = 4cose, तब4sine-3cose+ 2 का मान ज्ञात कीजिए।

Answer»

ANSWER:

Kyoooooioooooooooooo

Step-by-step EXPLANATION:

oooooooooooooooooo

Discussion

11.	Find the remainder when x¹⁹⁹⁹ is divided by x²-1.Plz solve
Answer» ANSWER: X²-1=0(to FIND its ZERO) X=1 Putting this in x^1999 So REMAINDER = 1 Step-by-step EXPLANATION:

11.

Find the remainder when x¹⁹⁹⁹ is divided by x²-1.Plz solve

Answer»

ANSWER:

X²-1=0(to FIND its ZERO)

X=1

Putting this in x^1999

So REMAINDER = 1

Step-by-step EXPLANATION:

Discussion

12.	factorise 8 x cube + 64 y cube + 48 x square Y + 96 y square then find the value of x + 2 Y is equal to 10
Answer» Answer: 64y CUBE ANS is ⅖ VALUE is 20 Step-by-step explanation: 2 +y =60

12.

factorise 8 x cube + 64 y cube + 48 x square Y + 96 y square then find the value of x + 2 Y is equal to 10

Answer»

Answer:

64y CUBE ANS is ⅖ VALUE is 20

Step-by-step explanation:

2 +y =60

Discussion

13.	Divide (2x3 - 12x2+ 16x)/(x-2)(x-4)
Answer» Answer: 2x • (x - 4)2 Step-by-step explanation: Step by Step Solution More Icon Reformatting the input : CHANGES made to your input should not affect the solution: (1): "x2" was replaced by "x^2". 1 more similar replacement(s). STEP 1 : Equation at the end of step 1 (((2•(x3))-(22•3x2))+16x) —————————————————————————•(x-4) (x-2) STEP 2 : Equation at the end of step 2 : ((2x3 - (22•3x2)) + 16x) ———————————————————————— • (x - 4) (x - 2) STEP 3 : 2x3 - 12x2 + 16x Simplify ———————————————— x - 2 STEP 4 : Pulling out like terms 4.1 Pull out like factors : 2x3 - 12x2 + 16x = 2x • (x2 - 6x + 8) Trying to factor by splitting the middle term 4.2 Factoring x2 - 6x + 8 The FIRST term is, x2 its coefficient is 1 . The middle term is, -6x its coefficient is -6 . The last term, "the constant", is +8 Step-1 : Multiply the coefficient of the first term by the constant 1 • 8 = 8 Step-2 : Find two factors of 8 whose sum equals the coefficient of the middle term, which is -6 . -8 + -1 = -9 -4 + -2 = -6 That's it Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and -2 x2 - 4x - 2x - 8 Step-4 : Add up the first 2 terms, pulling out like factors : x • (x-4) Add up the last 2 terms, pulling out common factors : 2 • (x-4) Step-5 : Add up the four terms of step 4 : (x-2) • (x-4) Which is the desired factorization Canceling Out : 4.3 Cancel out (x-2) which appears on both sides of the fraction line. Equation at the end of step 4 : 2x • (x - 4) • (x - 4) STEP 5 : Multiplying EXPONENTIAL Expressions: 5.1 Multiply (x-4) by (x-4) The rule says : To multiply exponential expressions which have the same base, add up their exponents. In our CASE, the common base is (x-4) and the exponents are : 1 , as (x-4) is the same number as (x-4)1 and 1 , as (x-4) is the same number as (x-4)1 The product is therefore, (x-4)(1+1) = (x-4)2 Final result : 2x • (x - 4)2 MARK AS BRAINLIEST

13.

Divide (2x3 - 12x2+ 16x)/(x-2)(x-4)

Answer»

Answer:

2x • (x - 4)2

Step-by-step explanation:

Step by Step Solution

More Icon

Reformatting the input :

CHANGES made to your input should not affect the solution:

(1): "x2" was replaced by "x^2". 1 more similar replacement(s).

STEP

1

:

Equation at the end of step 1

(((2•(x3))-(22•3x2))+16x)

—————————————————————————•(x-4)

(x-2)

STEP

2

:

Equation at the end of step

2

:

((2x3 - (22•3x2)) + 16x)

———————————————————————— • (x - 4)

(x - 2)

STEP

3

:

2x3 - 12x2 + 16x

Simplify ————————————————

x - 2

STEP

4

:

Pulling out like terms

4.1 Pull out like factors :

2x3 - 12x2 + 16x = 2x • (x2 - 6x + 8)

Trying to factor by splitting the middle term

4.2 Factoring x2 - 6x + 8

The FIRST term is, x2 its coefficient is 1 .

The middle term is, -6x its coefficient is -6 .

The last term, "the constant", is +8

Step-1 : Multiply the coefficient of the first term by the constant 1 • 8 = 8

Step-2 : Find two factors of 8 whose sum equals the coefficient of the middle term, which is -6 .

-8 + -1 = -9

-4 + -2 = -6 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and -2

x2 - 4x - 2x - 8

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x-4)

Add up the last 2 terms, pulling out common factors :

2 • (x-4)

Step-5 : Add up the four terms of step 4 :

(x-2) • (x-4)

Which is the desired factorization

Canceling Out :

4.3 Cancel out (x-2) which appears on both sides of the fraction line.

Equation at the end of step

4

:

2x • (x - 4) • (x - 4)

STEP

5

:

Multiplying EXPONENTIAL Expressions:

5.1 Multiply (x-4) by (x-4)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our CASE, the common base is (x-4) and the exponents are :

1 , as (x-4) is the same number as (x-4)1

and 1 , as (x-4) is the same number as (x-4)1

The product is therefore, (x-4)(1+1) = (x-4)2

Final result :

2x • (x - 4)2

MARK AS BRAINLIEST

Discussion

14.	If 432base 6-114base 6=z base 3
Answer» Step-by-step EXPLANATION: this MEET is FRIENDSHIP pug-ehak-qya

14.

If 432base 6-114base 6=z base 3

Answer»

Step-by-step EXPLANATION:

this MEET is FRIENDSHIP pug-ehak-qya

Discussion

15.	68 x 93 x 25333 X 43 X156
Answer» Drawing Daisy Wheel 5 3 3 А B 2 2 2 1 1 1 1 This is also called a six point geometry. Six rose petals, seed of LIFE, Dutch Hexagram, thunder ma Web Links Start with radius AB on two perpendicular lines. Place the COMPASS POINTER at B and draw the circle with radius AB. NEXT, place your compass at point 1 and using the same radius AB SW

15.

68 x 93 x 25333 X 43 X156

Answer»

Drawing Daisy Wheel

5

3

А

B

2

1

This is also called a six point geometry. Six rose petals, seed of LIFE, Dutch Hexagram, thunder ma

Web Links

Start with radius AB on two perpendicular lines. Place the COMPASS POINTER at B and draw the

circle with radius AB. NEXT, place your compass at point 1 and using the same radius AB SW

Discussion

16.	12. In ABC, D and E are the mid points of sides of AC and BC. If AB = 8.2 cm, find thelength of DE.
Answer» Answer: As D and E are the mid points of side AB and AC respectively so by Mid point theorem The LINE segment connecting the MIDPOINTS of two sides of a triangle is parallel to the THIRD side and is congruent to one half of the third side. DE=1/2BC DE=8.2/2 DE=4.1cm

16.

12. In ABC, D and E are the mid points of sides of AC and BC. If AB = 8.2 cm, find thelength of DE.

Answer»

Answer:

As D and E are the mid points of side AB and AC respectively

so by Mid point theorem

The LINE segment connecting the MIDPOINTS of two sides of a triangle is parallel to the THIRD side and is congruent to one half of the third side.

DE=1/2BC

DE=8.2/2

DE=4.1cm

Discussion

17.	(iii) First 10 multiples of 3.
Answer» Answer: 3,6,9,12 ,30 Step-by-step EXPLANATION: MARK me as a BRAINLIST

17.

(iii) First 10 multiples of 3.

Answer»

Answer:

3,6,9,12 ,30

Step-by-step EXPLANATION:

MARK me as a BRAINLIST

Discussion

18.	Solve the following of 5/6 - 2/3
Answer» ANSWER = 5/6 - 2/3 × 2/2 = 5/6- 4/6 = 1/6

18.

Solve the following of 5/6 - 2/3

Answer»

ANSWER

= 5/6 - 2/3 × 2/2

= 5/6- 4/6

= 1/6

Discussion

19.	What is rhs congruece rule
Answer» Answer: I think your QUESTION is wrong PLS WRITE correct question full question

19.

What is rhs congruece rule

Answer»

Answer:

I think your QUESTION is wrong PLS WRITE correct question

full question

Discussion

20.	State corresponding angles axiom.
Answer» Corresponding Angle Axiom: If a TRANSVERSAL intersects two parallel lines, then each pair of corresponding angles are EQUAL. Converse: If a transversal intersects two parallel lines such that a pair of corresponding angles is equal then the two lines are parallel. Mark my ANSWER as Brainlist please

20.

State corresponding angles axiom.

Answer»

Corresponding Angle Axiom: If a TRANSVERSAL intersects two parallel lines, then each pair of corresponding angles are EQUAL. Converse: If a transversal intersects two parallel lines such that a pair of corresponding angles is equal then the two lines are parallel.

Mark my ANSWER as Brainlist please

Discussion

21.	SRNO- 32CO2 - Solve the following Sub question decide whetherIndividual have to poyincome TaxIndividualAge taxable incomemiss Reena2522120,000ulo 2 to Mr Raju22185,000My Kailash67 23 100,000Mr Vinod4624180,0005 Morvilhou25, 82, ooorrassusjaya8) 2450.00034.496
Answer» ANSWER: I didn't UNDERSTAND Step-by-step EXPLANATION: because it is in my further CLASSES

21.

SRNO- 32CO2 - Solve the following Sub question decide whetherIndividual have to poyincome TaxIndividualAge taxable incomemiss Reena2522120,000ulo 2 to Mr Raju22185,000My Kailash67 23 100,000Mr Vinod4624180,0005 Morvilhou25, 82, ooorrassusjaya8) 2450.00034.496

Answer»

ANSWER:

I didn't UNDERSTAND

Step-by-step EXPLANATION:

because it is in my further CLASSES

Discussion

22.	A piece of wire is 96cm long. What will be the length of each side,if the wire is bent to form a regular hexagon?a. 16 cmb. 6 cmc. 24 cmd. 14 cm
Answer» Given: A piece of wire is 96cm. ⠀ Need to find: What will be the length of each side? ⠀ ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ ⠀ ❒ Let US CONSIDER that, the length of each REQUIRED side be 'a'. ⠀ ◗ Let's Head to the QUESTION now: ⠀ $\underline{ \bf{\dag} \frak{ \: As \: we \: know \: that \: :}}$ ⠀ Perimeter of HEXAGON = Length of wire. ⠀ We know that, if we are given with the perimeter of hexagon, we have the required formula, that is, ⠀ $:\implies \textsf{Perimeter of hexagon = 6a}$ ⠀ Substituting all the given values in the formula, we get : ⠀ $\sf{:\implies \quad 96 = 6 \times a} \\ \\ \\ \sf{:\implies \quad 96 = 6a} \\ \\ \\ \sf{:\implies \quad 6a = 96} \\ \\ \\ \sf{:\implies \quad a = \cancel{\frac{96}{6} }} \\ \\ \\ \sf{:\implies \quad \boxed{ \frak{ \pink{a = 16}}}}$ ⠀ $\sf{\therefore{\underline{The \: length \: of \: each \: required \: side \: is \: \textsf{\textbf{16 cm.}}}}}$

22.

A piece of wire is 96cm long. What will be the length of each side,if the wire is bent to form a regular hexagon?a. 16 cmb. 6 cmc. 24 cmd. 14 cm

Answer»

Given: A piece of wire is 96cm.

⠀

Need to find: What will be the length of each side?

⠀

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀

❒ Let US CONSIDER that, the length of each REQUIRED side be 'a'.

⠀

◗ Let's Head to the QUESTION now:

⠀

$\underline{ \bf{\dag} \frak{ \: As \: we \: know \: that \: :}}$

⠀

Perimeter of HEXAGON = Length of wire.

⠀

We know that, if we are given with the perimeter of hexagon, we have the required formula, that is,

⠀

$:\implies \textsf{Perimeter of hexagon = 6a}$

⠀

Substituting all the given values in the formula, we get :

⠀

$\sf{:\implies \quad 96 = 6 \times a} \\ \\ \\ \sf{:\implies \quad 96 = 6a} \\ \\ \\ \sf{:\implies \quad 6a = 96} \\ \\ \\ \sf{:\implies \quad a = \cancel{\frac{96}{6} }} \\ \\ \\ \sf{:\implies \quad \boxed{ \frak{ \pink{a = 16}}}}$

⠀

$\sf{\therefore{\underline{The \: length \: of \: each \: required \: side \: is \: \textsf{\textbf{16 cm.}}}}}$

Discussion

23.	State true/false There is no perfect cube which end with 8.
Answer» ANSWER: YES there is no PERFECT CUBE

23.

State true/false There is no perfect cube which end with 8.

Answer»

ANSWER:

YES there is no PERFECT CUBE

Discussion

24.	Solve the system of equationsusing Cramers rule3x + 24 =10- 6x + 4y = 4
Answer» Answer: XAM KI tu no ONE 1⃣ can PROPOSE you have to you☄ and the rest is HISTORY teacher‍‍ is a good day of my life is not

24.

Solve the system of equationsusing Cramers rule3x + 24 =10- 6x + 4y = 4

Answer»

Answer:

XAM KI tu no ONE 1⃣ can PROPOSE you have to you☄ and the rest is HISTORY teacher‍‍ is a good day of my life is not

Discussion

25.	Q) Four times the area of curved surface of cylinderis equal to 6 tinmes the sum of area of its bases. If itsheight is 12cm, find the curved surface area
Answer» ANSWER: CURVED SURFACE AREA= 2πrh sq.cm Area of the BASE = πr² sq.cm. Sum area of the base= πr²+πr²= 2πr²sq.cm A.T.Q 4(2πrh)= 6×2πr² 8πrh=12πr² 2h=3r 2×12=3r r=8cm. Curved Surface Area=2πrh = 2×22/7×8×12 =4224/7 = 603.42 sq.cm

25.

Q) Four times the area of curved surface of cylinderis equal to 6 tinmes the sum of area of its bases. If itsheight is 12cm, find the curved surface area

Answer»

ANSWER:

CURVED SURFACE AREA= 2πrh sq.cm

Area of the BASE = πr² sq.cm.

Sum area of the base= πr²+πr²= 2πr²sq.cm

A.T.Q

4(2πrh)= 6×2πr²

8πrh=12πr²

2h=3r

2×12=3r

r=8cm.

Curved Surface Area=2πrh

= 2×22/7×8×12

=4224/7

= 603.42 sq.cm

Discussion

26.	Find the value of x:5²x+¹=5³×625
Answer» ANSWER: I don't know Step-by-step EXPLANATION: I. . . @. . . .. . . ..

26.

Find the value of x:5²x+¹=5³×625

Answer»

ANSWER:

I don't know

Step-by-step EXPLANATION:

I. .

.

@.

.

..

.

..

Discussion

27.	Median for groupd data
Answer» ANSWER: bahut badhiya QUESTION hai beta bahut aage JAO GE

27.

Median for groupd data

Answer»

ANSWER:

bahut badhiya QUESTION hai beta bahut aage JAO GE

Discussion

28.	The numerator of a fraction is 7 less than the denominator if 3 is added to the numerator and2 is added to the denominator the value of the fraction becomes 2/3 find the original fraction
Answer» Step-by-step EXPLANATION: Let the numerator be x and the DENOMINATOR be y Given that numerator of a fraction is 7 less than its denominator Therefore, x+7=y ------(1) Given that if the denominator increased by 9 and the numerator by 2 then the fraction is 2/3 Therefore, X+2/y+9 = 2/3 ⟹3x+6=2y+18 ⟹3x−2y=12 ⟹3x−2(x+7)=12 ⟹x−14=12 ⟹x=26 ⟹y=26+7=33 (from (1)) Therefore, the fraction is 26/33 hope it is HELPFUL to you

28.

The numerator of a fraction is 7 less than the denominator if 3 is added to the numerator and2 is added to the denominator the value of the fraction becomes 2/3 find the original fraction

Answer»

Step-by-step EXPLANATION:

Let the numerator be x and

the DENOMINATOR be y

Given that numerator of a fraction is 7 less than its denominator

Therefore, x+7=y ------(1)

Given that if the denominator increased by 9 and the numerator by 2 then the fraction is 2/3

Therefore, X+2/y+9 = 2/3

⟹3x+6=2y+18

⟹3x−2y=12

⟹3x−2(x+7)=12

⟹x−14=12

⟹x=26

⟹y=26+7=33 (from (1))

Therefore, the fraction is 26/33

hope it is HELPFUL to you

Discussion

29.	- CONVERT DECIMAL NUMBER INTO ITS EQUIVALENT BINARY NUMBERa. 15b. 322C. 150d. 140e. 68f. 96g. 48h. 455i. 255j. 12
Answer» we will divide 15 by 270 14 and 10 n-1 will COME the answer 322 we will divide by 2 and answer will come 1162 we will divide 150 answer will come 0.40 answer 10 will do it 6896 answer will come 0 divide 48 hours and become zero we will divide 455 answer will come 1882 55171 we will do it 12 answer will come 0

Discussion

30.	3x+4=5(x-2) please solve this as fast as u can
Answer» ANSWER: this is IMPORTANT QUESTION so do it yourself

30.

3x+4=5(x-2) please solve this as fast as u can

Answer»

ANSWER:

this is IMPORTANT QUESTION so do it yourself

Discussion

31.	The mean weight of a group of 15 students is 62.5kg. If a new student is added and he weighs 75kg what is the new mean?
Answer» $\huge\boxed{\fcolorbox{purple}{ink}{Answer}}$ Original mean = 62.5 kg Original NUMBER of students = 15 students We KNOW that, ${\sf{Mean\: =\: {\dfrac{Sum\: of\: <klux>WEIGHT</klux>\: of\: students}{Total\: no\: of\: students}}}}$ ${\sf{\implies\: 62.5\: =\: {\dfrac{Sum\: of\: weight\: of\: students}{15}}}}$ ${\sf{\implies\: Sum\: of\: weight\: of\: students\: =\: 62.5\: \times\: 15}}$ = 937.5 kg New sum of weight = 937.5kg + 75kg = 1012.5 kg New no. of students = 15 + 1 = 16 ${\sf{New\:Mean\: =\: {\dfrac{New\:sum\: of\: weight}{New\:total\: no\: of\: students}}}}$ ${\sf{\implies\: New\:Mean\: =\: {\dfrac{1012.5}{16}}}}$ = 63.28 kg ${\bold{\boxed{\blue{New\: Mean\: =\: 63.28 kg}}}}$

31.

The mean weight of a group of 15 students is 62.5kg. If a new student is added and he weighs 75kg what is the new mean?

Answer»

$\huge\boxed{\fcolorbox{purple}{ink}{Answer}}$

Original mean = 62.5 kg

Original NUMBER of students = 15 students

We KNOW that,

${\sf{Mean\: =\: {\dfrac{Sum\: of\: <klux>WEIGHT</klux>\: of\: students}{Total\: no\: of\: students}}}}$

${\sf{\implies\: 62.5\: =\: {\dfrac{Sum\: of\: weight\: of\: students}{15}}}}$

${\sf{\implies\: Sum\: of\: weight\: of\: students\: =\: 62.5\: \times\: 15}}$

= 937.5 kg

New sum of weight = 937.5kg + 75kg = 1012.5 kg

New no. of students = 15 + 1 = 16

${\sf{New\:Mean\: =\: {\dfrac{New\:sum\: of\: weight}{New\:total\: no\: of\: students}}}}$

${\sf{\implies\: New\:Mean\: =\: {\dfrac{1012.5}{16}}}}$

= 63.28 kg

${\bold{\boxed{\blue{New\: Mean\: =\: 63.28 kg}}}}$

Discussion

32.	(2x - y) dx + y dy = 0solve Plzen
Answer» ANSWER: x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2× x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×

32.

(2x - y) dx + y dy = 0solve Plzen

Answer»

ANSWER:

x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×x7/2×

Discussion

33.	What should be taken away from 5x2 – 6y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20?]
Answer» ANSWER: hdydhdddhdfjfrj to the FOLLOWING QUADRATIC EQUATIONS and Step-by-step EXPLANATION: eusuf to the following quadratic equations

33.

What should be taken away from 5x2 – 6y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20?]

Answer»

ANSWER:

hdydhdddhdfjfrj to the FOLLOWING QUADRATIC EQUATIONS and

Step-by-step EXPLANATION:

eusuf to the following quadratic equations

Discussion

34.	Polynomial of digree 2 is called cubic polynomial
Answer» ANSWER: The given STATEMENT is false. $\rule{300}{<klux>2</klux>}$ EXPLANATION: The polynomial of degree of 2 is called a quadratic polynomial. But ACCORDING to the given equation it is cubic polynomial. Hence the given statement is false. $\rule{300}{2}$ Constant polynomial: The polynomial of degree 0 is called constant polynomial. For example→ 2x⁰+6 $\rule{300}{2}$ Linear polynomial: The polynomial of degree 1 is called Linear polynomial. For example→ 2x¹+5 $\rule{300}{2}$ Quadratic polynomial: The polynomial with degree 2 is called a quadratic polynomial. For example→ 5x²+7 $\rule{300}{2}$ Cubic polynomial: The polynomial with degree 3 is called cubic polynomial. For example→ 7x³+9 $\rule{300}{2}$ ADDITIONAL INFORMATION: What is a degree in a polynomial? The highest power in a polynomial is called it's degree. For example: 3x⁴+6x³+5x²+x Here in this polynomial, it's degree is 4 because it is the highest power. $\rule{300}{2}$

34.

Polynomial of digree 2 is called cubic polynomial

Answer»

ANSWER:

The given STATEMENT is false.

$\rule{300}{<klux>2</klux>}$

EXPLANATION:

The polynomial of degree of 2 is called a quadratic polynomial. But ACCORDING to the given equation it is cubic polynomial. Hence the given statement is false.

$\rule{300}{2}$

Constant polynomial:

The polynomial of degree 0 is called constant polynomial.

For example→ 2x⁰+6

$\rule{300}{2}$

Linear polynomial:

The polynomial of degree 1 is called Linear polynomial.

For example→ 2x¹+5

$\rule{300}{2}$

Quadratic polynomial:

The polynomial with degree 2 is called a quadratic polynomial.

For example→ 5x²+7

$\rule{300}{2}$

Cubic polynomial:

The polynomial with degree 3 is called cubic polynomial.

For example→ 7x³+9

$\rule{300}{2}$

ADDITIONAL INFORMATION:

What is a degree in a polynomial?

The highest power in a polynomial is called it's degree.

For example:

3x⁴+6x³+5x²+x

Here in this polynomial, it's degree is 4 because it is the highest power.

$\rule{300}{2}$

Discussion

35.	Find the volume of a cubical box whose surface area is 484cm²
Answer» $\huge\purple{\mathbb{★Answer★}}$ EDGE = a Surface AREA = 6a² = 486cm² 6a² = 486cm² ${a}^{2} = \frac{486}{2 {}^{} }$ a² = 81 a = √81 a = 9cm Edge = a = 9cm Volume = a³ = 9³ = 729³ Please GIVE me a ❤️☺️ :)

35.

Find the volume of a cubical box whose surface area is 484cm²

Answer»

$\huge\purple{\mathbb{★Answer★}}$

EDGE = a

Surface AREA = 6a² = 486cm²

6a² = 486cm²

${a}^{2} = \frac{486}{2 {}^{} }$

a² = 81

a = √81

a = 9cm

Edge = a = 9cm

Volume = a³ = 9³ = 729³

Please GIVE me a ❤️☺️ :)

Discussion

36.	SECTION - BWhat is the multiplicative inverse of[ 21/5-(-1/3)
Answer» ANSWER: ANSWER IS 15/68 Step-by-step EXPLANATION: MARK ME AS BRAINLIEST

36.

SECTION - BWhat is the multiplicative inverse of[ 21/5-(-1/3)

Answer»

ANSWER:

ANSWER IS 15/68

Step-by-step EXPLANATION:

MARK ME AS BRAINLIEST

Discussion

37.	Which of the following is the value of (x+1/2)2?
Answer» ANSWER: the VALUE of (x+1/2)2 2x+2/2=0. x+1=0 x=-1 hope U LIKE

37.

Which of the following is the value of (x+1/2)2?

Answer»

ANSWER:

the VALUE of (x+1/2)2

2x+2/2=0.

x+1=0

x=-1

hope U LIKE

Discussion

38.	Solve the system of equationsusing Csamens rule3x + 2y = 10- 6x + 4y = 4need answer
Answer» ANSWER: PM that 3IS 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3person PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwjPM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3person 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3person Step-by-step EXPLANATION: PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3person PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3personHenderson is the only 33PM that 3is 3for 3to 63363to the 3home ewhwj 3person

38.