332354 + Interview Questions in Secondary School in Math

1.	Solve a) x/4=7b)9y=0.64c)5/10.x=67solve please follow me
Answer» HOPE...it will HELP you...

Discussion

2.	24x^2 – 15x by spiltting middle term
Answer» ANSWER: 3x(8X- 5) is the answer

2.

24x^2 – 15x by spiltting middle term

Answer»

ANSWER:

3x(8X- 5) is the answer

Discussion

3.	9. Sharad is facing towards east. He turns by 135°anti-clockwise then 270° in clockwise direction.Finally he takes right turn. The direction in whichhe is facing now, is
Answer» ANSWER: SOUTH- East Step-by-step explanation: 270°-135°=135° in CLOCKWISE direction. He is FACING in South-East

3.

9. Sharad is facing towards east. He turns by 135°anti-clockwise then 270° in clockwise direction.Finally he takes right turn. The direction in whichhe is facing now, is

Answer»

ANSWER:

SOUTH- East

Step-by-step explanation:

270°-135°=135° in CLOCKWISE direction.

He is FACING in South-East

Discussion

4.	Round of the following numbers to the nearest thousand 7654
Answer» Answer: 8000.. .......... ...

4.

Round of the following numbers to the nearest thousand 7654

Answer»

Answer:

8000.. .......... ...

Discussion

5.	What is isosceles trapezium ????
Answer» Holå mate❤ ANSWER✍ ✨the QUADRILATERAL of which ONE pair of opposite SIDES is paralleland other pair of opposite sides is UNPARALLEL and equal i.e.,the two TRANSVERSAL sides are equal is called isosceles trapezium .✨ hope it's help you✌

5.

What is isosceles trapezium ????

Answer»

Holå mate❤

ANSWER✍

✨the QUADRILATERAL of which ONE pair of opposite SIDES is paralleland other pair of opposite sides is UNPARALLEL and equal i.e.,the two TRANSVERSAL sides are equal is called isosceles trapezium .✨

hope it's help you✌

Discussion

6.	Find the value of tan 55°. tan 45º. tan 35º.
Answer» Step-by-step explanation: USE the rule of COMPLEMENTARITY of angles....and u GET UR RESULT.... hope it helps...

6.

Find the value of tan 55°. tan 45º. tan 35º.

Answer»

Step-by-step explanation:

USE the rule of COMPLEMENTARITY of angles....and u GET UR RESULT....

hope it helps...

Discussion

7.	P and Q are the mid points of the sides AB and BC respectively of the triangle ABC, right-angled at B, then?
Answer» ANSWER: PQB is triangle Step-by-step EXPLANATION:

7.

P and Q are the mid points of the sides AB and BC respectively of the triangle ABC, right-angled at B, then?

Answer»

ANSWER:

PQB is triangle

Step-by-step EXPLANATION:

Discussion

8.	Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzxzzzxxxxxxxfehshdhdudjudhr7rheud12 help question
Answer» HEY MATE! PLEASE complete the question before posting it....

Discussion

9.	the adjacent sides of parallelogram are 20 cm and 15 cm and length in the ratio between the corresponding altitude is
Answer» Hey Dear, ◆ Answer - 3 : 4 ● Explanation - Given that adjacent sides of PARALLELOGRAM are - X = 20 cm x = 20 cmy = 15 cm Let a & b be corresponding altitudes of the parallelogram. Area of parallelogram is CALCULATED as PRODUCT of side length with corresponding altitude. Area = x.a = y.b a/b = y/x a/b = 15/20 a/b = 3/4 Therefore, ratio of lengths of corresponding altitudes of parallelogram is 3 : 4. Thanks dear..

9.

the adjacent sides of parallelogram are 20 cm and 15 cm and length in the ratio between the corresponding altitude is

Answer»

Hey Dear,

◆ Answer -

3 : 4

● Explanation -

Given that adjacent sides of PARALLELOGRAM are -

X = 20 cm

x = 20 cmy = 15 cm

Let a & b be corresponding altitudes of the parallelogram.

Area of parallelogram is CALCULATED as PRODUCT of side length with corresponding altitude.

Area = x.a = y.b

a/b = y/x

a/b = 15/20

a/b = 3/4

Therefore, ratio of lengths of corresponding altitudes of parallelogram is 3 : 4.

Thanks dear..

Discussion

10.	. Milk is sold at Rs.37 per litre. Find the cost of 6 litres of milk.
Answer» $\Mathcal{\huge\green\star}{\red{\huge{Heya Mate}}}{\huge\green\star}$ QUES: Milk is SOLD at ₹ 37/ LITRE. Find the cost of 6 litres of milk. Ans: Cost of 1 litre milk➡️₹37 We need to find the cost of 6 litres, Hence, (cost of 1 litre milk × 6 ) is the cost of 6 litres of milk. Cost of 6 litres milk ➡️ 6 × 37 ➡️₹222 $\huge\mathcal\green{Hope\: it\:helps}$ Life ☯️ Love ❤️ PEACE ☮️

10.

. Milk is sold at Rs.37 per litre. Find the cost of 6 litres of milk.

Answer»

$\Mathcal{\huge\green\star}{\red{\huge{Heya Mate}}}{\huge\green\star}$

QUES: Milk is SOLD at ₹ 37/ LITRE.

Find the cost of 6 litres of milk.

Ans:

Cost of 1 litre milk➡️₹37

We need to find the cost of 6 litres,

Hence, (cost of 1 litre milk × 6 ) is the cost of 6 litres of milk.

Cost of 6 litres milk ➡️ 6 × 37

➡️₹222

$\huge\mathcal\green{Hope\: it\:helps}$

Life ☯️ Love ❤️ PEACE ☮️

Discussion

11.	4. Simplify this please VV22
Answer» $\mathcal{\huge\red\star}{\red{\huge{Heymate}}}{\huge\red\star}$ ✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶ $\mathcal{\green{\huge{Answer}}}$ The SOLUTION is as follows $\sqrt[4]{ \sqrt[3]{ {2}^{2} } } \\ {2}^{2 \times \frac{1}{3} \times \frac{1}{4} } \\ {2}^{ \frac{1}{3} \times \frac{1}{2} } \\ 2 {}^{ \frac{1}{6} } \\ \sqrt[6]{2}$ ❤Thanks for ASKING $\underline\bold{ANJALI}$ ✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶ $\mathcal{\red{\huge {Hope\:it\:helps}}}$ ⍖━━━━━━━━━━⍖ $\mathbb{\huge{\blue{R-K}}}$ $\mathcal{\huge{\green{Plz_-Follow_-Me}}}$

11.

4. Simplify this please VV22

Answer»

$\mathcal{\huge\red\star}{\red{\huge{Heymate}}}{\huge\red\star}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\mathcal{\green{\huge{Answer}}}$

The SOLUTION is as follows

$\sqrt[4]{ \sqrt[3]{ {2}^{2} } } \\ {2}^{2 \times \frac{1}{3} \times \frac{1}{4} } \\ {2}^{ \frac{1}{3} \times \frac{1}{2} } \\ 2 {}^{ \frac{1}{6} } \\ \sqrt[6]{2}$

❤Thanks for ASKING $\underline\bold{ANJALI}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\mathcal{\red{\huge {Hope\:it\:helps}}}$

⍖━━━━━━━━━━⍖

$\mathbb{\huge{\blue{R-K}}}$

$\mathcal{\huge{\green{Plz_-Follow_-Me}}}$

Discussion

12.	9/8+3/5 of (3/4+3/5)
Answer» Answer: 27/2069/40 Step-by-step explanation:* TAKE LCM and then MULTIPLY the whole solved brackets

12.

9/8+3/5 of (3/4+3/5)

Answer»

Answer:

27/20*69/40

Step-by-step explanation:

TAKE LCM and then MULTIPLY the whole solved brackets

Discussion

13.	Find the amount andcompound interest on rs. 2000at 10%for2and half years
Answer» <P>ANSWER: A = P (1 + R / 100)no of YEARS A = 2000 11/10 * 11/10 A = 20 * 121 amount = 2420 CI = 420

13.

Find the amount andcompound interest on rs. 2000at 10%for2and half years

Answer»

<P>ANSWER:

A = P (1 + R / 100)*no of YEARS

A = 2000 * 11/10 * 11/10

A = 20 * 121

amount = 2420

CI = 420

Discussion

14.	Sum of even numbers till 300
Answer» Answer: 22,650 Step-by-step explanation: Given : First TERM ( a ) = 2 LAST term ( t_n ) = 300 Common difference ( d ) = 2 We have term formula t_n = a + ( N - 1 ) d 300 = 2 + ( n -1 ) 2 2 ( n - 1 ) = 298 n - 1 = 149 n = 150 So numbers of terms is 150 Now we have SUM formula S_n = n / 2 ( a + l ) where l is last term. $\displaystyle \text{$S_{150}=\frac{150}{2}(2+300) $}\\\\\displaystyle \text{$S_{150}=\frac{150}{2}(302) $}\\\\\displaystyle \text{$S_{150}=75\times302$}\\\\\displaystyle \text{$S_{150}=22,650$}$ Thus we get answer.

14.

Sum of even numbers till 300

Answer»

Answer:

22,650

Step-by-step explanation:

Given :

First TERM ( a ) = 2

LAST term ( t_n ) = 300

Common difference ( d ) = 2

We have term formula

t_n = a + ( N - 1 ) d

300 = 2 + ( n -1 ) 2

2 ( n - 1 ) = 298

n - 1 = 149

n = 150

So numbers of terms is 150

Now we have SUM formula

S_n = n / 2 ( a + l )

where l is last term.

$\displaystyle \text{$S_{150}=\frac{150}{2}(2+300) $}\\\\\displaystyle \text{$S_{150}=\frac{150}{2}(302) $}\\\\\displaystyle \text{$S_{150}=75\times302$}\\\\\displaystyle \text{$S_{150}=22,650$}$

Thus we get answer.

Discussion

15.	Choose the correct option and give a reason in support of youranswer.)XCIII + III =(a)CXVI(b) VXCI(c) XCVI
Answer» ANSWER: The answer is (C) part. Please MARK me as BRAINLIEST

15.

Choose the correct option and give a reason in support of youranswer.)XCIII + III =(a)CXVI(b) VXCI(c) XCVI

Answer»

ANSWER:

The answer is (C) part.

Please MARK me as BRAINLIEST

Discussion

16.	The product of two numbers is 3072, if the l.c.m of the numbers is 192, find the h.c.f.
Answer» Answer: HCF will be 16 Step-by-step EXPLANATION: since HCF X LCM = PRODUCT of numbers hence HCF = 3072/192 = 16

16.

The product of two numbers is 3072, if the l.c.m of the numbers is 192, find the h.c.f.

Answer»

Answer: HCF will be 16

Step-by-step EXPLANATION:

since HCF X LCM = PRODUCT of numbers

hence HCF = 3072/192 = 16

Discussion

17.	unit digit of two digit number of four times its digit are reversed it is found that the resulting number is 54 than the original number find the number
Answer» Answer: udididjjd FU hsjsksikdnxhjsoaowij u PLZ mark it brain LIST but help me write a LETTER to editor plz

17.

unit digit of two digit number of four times its digit are reversed it is found that the resulting number is 54 than the original number find the number

Answer»

Answer:

udididjjd FU hsjsksikdnxhjsoaowij u PLZ mark it brain LIST but help me write a LETTER to editor plz

Discussion

18.	Find your age:-write expressions in one variable comparing your present age with at least three of your family members.-write the expressions for the comparison of ages before 5 years and after 5 years.-form a linear equation using these expressions and solve each of them to get your present age.
Answer» don't know about this answer FIND it out by UR selves y R U DEPENDING on others

18.

Find your age:-write expressions in one variable comparing your present age with at least three of your family members.-write the expressions for the comparison of ages before 5 years and after 5 years.-form a linear equation using these expressions and solve each of them to get your present age.

Answer»

don't know about this answer FIND it out by UR selves y R U DEPENDING on others

Discussion

19.	The zeroes of the polynomial 2-procal of its zeroes.
Answer» Step-by-step EXPLANATION: WRITE FULL QUESTION it is INCOMPLETE

19.

The zeroes of the polynomial 2-procal of its zeroes.

Answer»

Step-by-step EXPLANATION:

WRITE FULL QUESTION it is INCOMPLETE

Discussion

20.	A shopkeeper offers his customers seccesive discounts of 10% and 5% respectively. IF the marked price of an article be Rs 400, find its selling price.
Answer» ANSWER: selling PRICE of the article is ₹342 Step-by-step explanation: MARKED price=Rs 400 first discount= 10%= ₹10/100×400=₹40 therefore,present s.p. =₹400-40=₹360 second discount=5%=₹5/100×360=₹18 therefore,s.p. of the article=₹360-18=₹342

20.

A shopkeeper offers his customers seccesive discounts of 10% and 5% respectively. IF the marked price of an article be Rs 400, find its selling price.

Answer»

ANSWER:

selling PRICE of the article is ₹342

Step-by-step explanation:

MARKED price=Rs 400

first discount= 10%= ₹10/100×400=₹40

therefore,present s.p. =₹400-40=₹360

second discount=5%=₹5/100×360=₹18

therefore,s.p. of the article=₹360-18=₹342

Discussion

21.	Can u say me the meaning of this symbol ^
Answer» ANSWER: it MEANS to the POWER............

21.

Can u say me the meaning of this symbol ^

Answer»

ANSWER:

it MEANS to the POWER............

Discussion

22.	The value of 0.3bar as a rational no. is?
Answer» ANSWER: YES it is a RATIONAL NUMBER

22.

The value of 0.3bar as a rational no. is?

Answer»

ANSWER:

YES it is a RATIONAL NUMBER

Discussion

23.	Answer the question in attachment
Answer» $\bf{\large{\underline{\underline{Question:-}}}}$ $\large{ \sf Simplify \: \dfrac{ {9}^{\frac{1}{3}} \times {27}^{ - \frac{1}{2} } }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ $\bf{\large{\underline{\underline{Solution:-}}}}$ $\large{\sf \dfrac{ {9}^{\frac{1}{3}} \times {27}^{ - \frac{1}{2} } }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ $\large{ \sf = \dfrac{ {({3}^{2})}^{\frac{1}{3}} \times {( {3}^{3})}^{ - \frac{1}{2} } }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ [In NUMERATOR $\sf {9}^{\frac{1}{3}} = {({3}^{2})}^{\frac{1}{3}}$ because 9 can be written as 3² and $\sf {27}^{-\frac{1}{2}} = {({3}^{3})}^{-\frac{1}{2}}$ because 27 can be written as 3³] $\large{ \sf = \dfrac{ {(3)}^{\frac{1}{3} \times 2} \times {(3)}^{ - \frac{1}{2} \times3 } }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ [LAW of exponents used in numerator $\sf (a^m)^n = a^{mn}$ ] $\large{ \sf = \dfrac{ {(3)}^{\frac{2}{3}} \times {(3)}^{ - \frac{3}{2}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ $\large{ \sf = \dfrac{ {(3)}^{\frac{2}{3}+ ( - \frac{3}{2})} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ [Law of exponents used in numerator $\sf a^m \times a^n = a^{m + n}$ ] $\large{ \sf = \dfrac{ {(3)}^{\frac{2}{3}- \frac{3}{2}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ $\large{ \sf = \dfrac{ {(3)}^{\frac{2 \times 2}{3 \times 2}- \frac{3 \times 3}{2 \times 3}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ $\large{ \sf = \dfrac{ {(3)}^{\frac{4}{6}- \frac{9}{6}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ $\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$ $\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{1}{8} + ( - \frac{2}{3})}} }$ [Law of exponents used in numerator $\sf a^m \times a^n = a^{m + n}$ ] $\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{1}{8}- \frac{2}{3}}} }$ $\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{1 \times 3}{8 \times 3}- \frac{2 \times 8}{3 \times 8}}} }$ $\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{3}{24}- \frac{16}{24}}} }$ $\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{- \frac{13}{24}}} }$ $\large{ \sf =(3)^{ - \frac{5}{6} - ( - \frac{13}{24}) } }$ [Law of EXPONENT used $\sf a^m \div a^n = a^{m-n}$ ] $\large{ \sf =(3)^{ - \frac{5}{6} + \frac{13}{24}} }$ $\large{ \sf =(3)^{ - \frac{5 \times 4}{6 \times 4} + \frac{13}{24}} }$ $\large{ \sf =(3)^{ - \frac{20}{24} + \frac{13}{24}} }$ $\large{ \sf =3^{ - \frac{7}{24}} }$ $\Huge{ \sf = \dfrac{1}{ {3}^{ \frac{7}{24} } } }$ [Law of exponent used $\sf a^{-m} = \dfrac{1}{a^m}$ ]

23.

Answer the question in attachment

Answer»

$\bf{\large{\underline{\underline{Question:-}}}}$

$\large{ \sf Simplify \: \dfrac{ {9}^{\frac{1}{3}} \times {27}^{ - \frac{1}{2} } }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

$\bf{\large{\underline{\underline{Solution:-}}}}$

$\large{\sf \dfrac{ {9}^{\frac{1}{3}} \times {27}^{ - \frac{1}{2} } }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

$\large{ \sf = \dfrac{ {({3}^{2})}^{\frac{1}{3}} \times {( {3}^{3})}^{ - \frac{1}{2} } }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

[In NUMERATOR $\sf {9}^{\frac{1}{3}} = {({3}^{2})}^{\frac{1}{3}}$ because 9 can be written as 3² and $\sf {27}^{-\frac{1}{2}} = {({3}^{3})}^{-\frac{1}{2}}$ because 27 can be written as 3³]

$\large{ \sf = \dfrac{ {(3)}^{\frac{1}{3} \times 2} \times {(3)}^{ - \frac{1}{2} \times3 } }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

[LAW of exponents used in numerator $\sf (a^m)^n = a^{mn}$ ]

$\large{ \sf = \dfrac{ {(3)}^{\frac{2}{3}} \times {(3)}^{ - \frac{3}{2}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

$\large{ \sf = \dfrac{ {(3)}^{\frac{2}{3}+ ( - \frac{3}{2})} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

[Law of exponents used in numerator $\sf a^m \times a^n = a^{m + n}$ ]

$\large{ \sf = \dfrac{ {(3)}^{\frac{2}{3}- \frac{3}{2}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

$\large{ \sf = \dfrac{ {(3)}^{\frac{2 \times 2}{3 \times 2}- \frac{3 \times 3}{2 \times 3}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

$\large{ \sf = \dfrac{ {(3)}^{\frac{4}{6}- \frac{9}{6}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

$\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{1}{8}} \times {3}^{ - \frac{2}{3} } } }$

$\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{1}{8} + ( - \frac{2}{3})}} }$

[Law of exponents used in numerator $\sf a^m \times a^n = a^{m + n}$ ]

$\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{1}{8}- \frac{2}{3}}} }$

$\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{1 \times 3}{8 \times 3}- \frac{2 \times 8}{3 \times 8}}} }$

$\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{ \frac{3}{24}- \frac{16}{24}}} }$

$\large{ \sf = \dfrac{ {(3)}^{- \frac{5}{6}} }{ {3}^{- \frac{13}{24}}} }$

$\large{ \sf =(3)^{ - \frac{5}{6} - ( - \frac{13}{24}) } }$

[Law of EXPONENT used $\sf a^m \div a^n = a^{m-n}$ ]

$\large{ \sf =(3)^{ - \frac{5}{6} + \frac{13}{24}} }$

$\large{ \sf =(3)^{ - \frac{5 \times 4}{6 \times 4} + \frac{13}{24}} }$

$\large{ \sf =(3)^{ - \frac{20}{24} + \frac{13}{24}} }$

$\large{ \sf =3^{ - \frac{7}{24}} }$

$\Huge{ \sf = \dfrac{1}{ {3}^{ \frac{7}{24} } } }$

[Law of exponent used $\sf a^{-m} = \dfrac{1}{a^m}$ ]

Discussion

24.	(A) factorise the following expression :-h) 36x^2 - 1/169i) 36(a + b)^2 - 49 (a - b)^2j) 24a^2 - 54b^2k) 3p^5 - 48p^3l) 63x^2 y^2 - 7 pls help me guys its urgent I need help plss solve these all questions
Answer» ANSWER: jdjdjjficicfiifufjj

24.

(A) factorise the following expression :-h) 36x^2 - 1/169i) 36(a + b)^2 - 49 (a - b)^2j) 24a^2 - 54b^2k) 3p^5 - 48p^3l) 63x^2 y^2 - 7 pls help me guys its urgent I need help plss solve these all questions

Answer»

ANSWER:

jdjdjjficicfiifufjj

Discussion

25.	6/x+2y+5/x-2y=-3 ; 3/x+2y+7/x-2y=-6
Answer» ANSWER: X = 0 and y = 1 Solution in ATTACHMENT :)

25.

6/x+2y+5/x-2y=-3 ; 3/x+2y+7/x-2y=-6

Answer»

ANSWER:

X = 0 and y = 1

Solution in ATTACHMENT :)

Discussion

26.	Express following in form of p/qi) 37.09158
Answer» ANSWER: 3709158/100000 BRO this is the answer

26.

Express following in form of p/qi) 37.09158

Answer»

ANSWER:

3709158/100000 BRO this is the answer

Discussion

27.	Two dice are rolled simultaneously. Find the probability that the sum is more than and equal to 10 or 12Experts plz answer step by step...
Answer» When two DICE rolled simultaneously TOTAL OUTCOMES = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} Total number of outcomes = 6 x 6 = 36 FAVOURABLE outcomes = {(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)} Number of favourable outcomes = 6 Probability = favourable outcomes/total outcomes Probability = 6/36 = 1/6

Discussion

28.	2/x-3/y=15;8/x+5/y=77
Answer» ANSWER: HOPE you got it so plse say THANKS and follow me alia JOSHI and MARK as brainlist plse

28.

2/x-3/y=15;8/x+5/y=77

Answer»

ANSWER:

HOPE you got it so plse say THANKS and follow me alia JOSHI and MARK as brainlist plse

Discussion

29.	Simple interest is added for what
Answer» Step-by-step EXPLANATION: FORMULA of SIMPLE interest I = PTR/100

29.

Simple interest is added for what

Answer»

Step-by-step EXPLANATION:

FORMULA of SIMPLE interest

I = PTR/100

Discussion

30.	Represent root 9.3 in a number line
Answer» Answer: STEP 1.draw a line and mark a point A on it step 2.mark a point on the line DRAWN in step 1 such that AB=9.3 cm step 3.mark a point on AB such that BC=1 cm step 4.find mid point of AC let the mid point be o step 5.taking o as a centre and oc=oa as radius draw a semicirle. also draw a line passing through B PERPENDICULAR to OB. suppose it cuts the semicirle at D step 6.taking B as centre and BD as radius draw an arc cutting OC produced at E. point E so obtained represents underoot 9.3

30.

Represent root 9.3 in a number line

Answer»

Answer:

STEP 1.draw a line and mark a point A on it

step 2.mark a point on the line DRAWN in step 1 such that AB=9.3 cm

step 3.mark a point on AB such that BC=1 cm

step 4.find mid point of AC let the mid point be o

step 5.taking o as a centre and oc=oa as radius draw a semicirle. also draw a line passing through B PERPENDICULAR to OB. suppose it cuts the semicirle at D

step 6.taking B as centre and BD as radius draw an arc cutting OC produced at E. point E so obtained represents underoot 9.3

Discussion

31.	72percent of 25 students are good in maths.how many are not good in maths ?
Answer» Answer: 7 Step-by-step explanation: total no of students = 25 now 72% of 25 =18 (72*25)/100 =18 so no of students INTERESTED in maths =18 total students =25 so students not interested = 25-18 =7

31.

72percent of 25 students are good in maths.how many are not good in maths ?

Answer»

Answer:

7

Step-by-step explanation:

total no of students = 25

now 72% of 25 =18

(72*25)/100

=18

so no of students INTERESTED in maths =18

total students =25

so students not interested

= 25-18

=7

Discussion

32.	Write the value of a^3+b^3+c^3, if a+b+c=0.
Answer» Answer: given, = a+b+c=0 -(1) =a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) -(2) PUT equation 1 in 2 =a^3+b^3+c^3-3abc=o (a^2+b^2+c^2-ab-bc-ca) =a^3+b^3+c^3-3abc=0 =a^3+b^3+c^3=3abc

32.

Write the value of a^3+b^3+c^3, if a+b+c=0.

Answer»

Answer:

given,

= a+b+c=0 -(1)

=a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) -(2)

PUT equation 1 in 2

=a^3+b^3+c^3-3abc=o (a^2+b^2+c^2-ab-bc-ca)

=a^3+b^3+c^3-3abc=0

=a^3+b^3+c^3=3abc

Discussion

33.	Find the product of (-320)×(-15)
Answer» Answer: 4850 I THINK it will HELP you

33.

Find the product of (-320)×(-15)

Answer»

Answer:

4850

I THINK it will HELP you

Discussion

34.	A sum of money is devided among 160 males and some females in the ratio 16: 21.Individually each male gets Rs.4 and female RS.3. What should be the number of females?
Answer» Answer: Here we have taken no.of FEMALES=x 160 * 4/x * 3=16/21 x=280 Hence the TOTAL no.of females are 280

34.

A sum of money is devided among 160 males and some females in the ratio 16: 21.Individually each male gets Rs.4 and female RS.3. What should be the number of females?

Answer»

Answer:

Here we have taken no.of FEMALES=x

160 * 4/x * 3=16/21

x=280

Hence the TOTAL no.of females are 280

Discussion

35.	If a,b are the zero of the polynomial (x)=x2p(x-1)-c,then(a+1) (b+1)
Answer» Answer: ns >> Extras >> STUDENT Questions Next similar questionQ) If alpha and beeta are zeroes of polynomial x2-p(x+1)+C such that (alpha+1) If alpha and beeta are zeroes of polynomial x2-p(x+1)+c such that (alpha+1)(beeta+1)=0. Then find the value of c. Homework help Donate A) Given that alpha and beta are the roots of the QUADRATIC EQUATION f(x) = x^2-p(x+1)-c = x^2-px-p-c = x^2 -px-(p+c), comparing with ax^2 + bx + c, we have, a =1 , b= -p & c= -(p+c) alpha+beta = -b/a = -(-p)/1 = p & alphabeta = c/a = -(p+c)/1 = -(p+c) Therefore, (Alpha + 1)(beta+1) = Alpha*beta + alpha + beta + 1 = -(p+c) + p + 1 = -p-c+p+1 = 1-c or c=1

35.

If a,b are the zero of the polynomial (x)=x2p(x-1)-c,then(a+1) (b+1)

Answer»

Answer:

ns >> Extras >> STUDENT Questions

Next similar questionQ)

If alpha and beeta are zeroes of polynomial x2-p(x+1)+C such that (alpha+1)

If alpha and beeta are zeroes of polynomial x2-p(x+1)+c such that (alpha+1)(beeta+1)=0. Then find the value of c.

Homework help Donate

A)

Given that alpha and beta are the roots of the QUADRATIC EQUATION f(x) = x^2-p(x+1)-c = x^2-px-p-c = x^2 -px-(p+c),

comparing with ax^2 + bx + c, we have, a =1 , b= -p & c= -(p+c)

alpha+beta = -b/a = -(-p)/1 = p

& alpha*beta = c/a = -(p+c)/1 = -(p+c)

Therefore, (Alpha + 1)*(beta+1)

= Alpha*beta + alpha + beta + 1

= -(p+c) + p + 1

= -p-c+p+1

= 1-c

or c=1

Discussion

36.	Prove that the characteristic of a field of zero is perfect
Answer» <P>ANSWER: characteristic 0fields have a very HANDY FEATURE every irreducible polynomial is characteristic 0 is separable. fields in characteristic p MAY or may not be have this feature A field Kis called perfect if every irreducible polynomial is K(x) is seperable.every field of characteristic of a field of zero is perfect

36.

Prove that the characteristic of a field of zero is perfect

Answer»

<P>ANSWER:

characteristic 0fields have a very HANDY FEATURE every irreducible polynomial is characteristic 0 is separable. fields in characteristic p MAY or may not be have this feature A field Kis called perfect if every irreducible polynomial is K(x) is seperable.every field of characteristic of a field of zero is perfect

Discussion

37.	Probability of non leap year having 52 mondays
Answer» Answer: probability of 52 Mondays in a normal YEAR is one because there is 52 Monday's in a year ALWAYS so it is CERTAIN EVENT WHOSE probability is 1

37.

Probability of non leap year having 52 mondays

Answer»

Answer:

probability of 52 Mondays in a normal YEAR is one because there is 52 Monday's in a year ALWAYS so it is CERTAIN EVENT WHOSE probability is 1

Discussion

38.	If 10x=0.2bar - 0.3bar find the value of x
Answer» ANSWER: x=-0.01bar Step-by-step explanation: 10x=0.2-0.3bar 10x=-0.1bar then x=0.01bar

38.

If 10x=0.2bar - 0.3bar find the value of x

Answer»

ANSWER:

x=-0.01bar

Step-by-step explanation:

10x=0.2-0.3bar

10x=-0.1bar

then x=0.01bar

Discussion

39.	If x minus 3 Y is equals to 3 and x y is equals to 18 find the value of x square + 9 y square
Answer» Answer: 117 Step-by-step explanation: square of x-3y is 9 x^2+9y^2 -6xy is 9 xy is 18 x^2+9y^2=9+108 x^2+9y^2=117 HOPE IT HELPS PLEASE MARK ME BRAINLIEST

39.

If x minus 3 Y is equals to 3 and x y is equals to 18 find the value of x square + 9 y square

Answer»

Answer:

117

Step-by-step explanation:

square of x-3y is 9

x^2+9y^2 -6xy is 9

xy is 18

x^2+9y^2=9+108

x^2+9y^2=117

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST

Discussion

40.	What is the remainder when """x=1/2 not X+1/2"""
Answer» ANSWER: doctoring calcinosis resubmission McDonald's xoxo lasso HARICOTS hasn't DU us up

40.

What is the remainder when """x=1/2 not X+1/2"""

Answer»

ANSWER:

doctoring calcinosis resubmission McDonald's xoxo lasso HARICOTS hasn't DU us up

Discussion

41.	How to do 3^3^8^0^5^6
Answer» 0 if we MULTIPLY any NUMBER with 0 we will GET 0

41.

How to do 3^3^8^0^5^6

Answer»

0

if we MULTIPLY any NUMBER with 0 we will GET 0

Discussion

42.	If 3root a + 3root b + 3root c =0 then find a+b+b whole cube
Answer» ANSWER: a+b+c=3abc=0. √3+√3+√3=√9.

42.

If 3root a + 3root b + 3root c =0 then find a+b+b whole cube

Answer»

ANSWER:

a+b+c=3abc=0.

√3+√3+√3=√9.

Discussion

43.	9÷2_6÷5÷2÷5×2+1 solution
Answer» ANSWER: Here is your answer friend. Step-by-step explanation: 5.26. PLEASE MARK as BRAINLIEST answer.

43.

9÷2_6÷5÷2÷5×2+1 solution

Answer»

ANSWER:

Here is your answer friend.

Step-by-step explanation:

5.26.

PLEASE MARK as BRAINLIEST answer.

Discussion

44.	Find the roots of quadratic equation :- ax²+bx+c=0 (a≠0)
Answer» Answer: here u go Step-by-step EXPLANATION: the roots of the GIVEN equation are $x = \frac{ - b + \sqrt[]{ {b}^{2} \: - 4ac } }{2a} \\ and \\x = \frac{ - b - \sqrt[]{ {b}^{2} \: - 4ac } }{2a} \:$ hope this helps u...... MARK me as BRAINLIEST...... FOLLOW ME......

44.

Find the roots of quadratic equation :- ax²+bx+c=0 (a≠0)

Answer»

Answer:

here u go

Step-by-step EXPLANATION:

the roots of the GIVEN equation are

$x = \frac{ - b + \sqrt[]{ {b}^{2} \: - 4ac } }{2a} \\ and \\x = \frac{ - b - \sqrt[]{ {b}^{2} \: - 4ac } }{2a} \:$

hope this helps u......

MARK me as BRAINLIEST......

FOLLOW ME......

Discussion

45.	If each side of a triangle is doubled, find the % increase in the area of the triangle
Answer» Answer: Area is INCREASED by 300%. Step-by-step explanation: If that triangle is a right angled triangle. = > Original area = 1 / 2 x base x height *{ From properties }* When sides are DOUBLED : = > NEW area = 1 / 2 x 2base x 2height = > New area = 2 x base x height Increase % : ( 2 x height x base ) / ( 1 / 2 x base x height ) x 100% = > 2 / ( 1 / 2 ) x 100% = > 4 x 100% = > 400% Hence the area is increased by 400%. If that triangle is not right angled triangle : Let the sides are 2a , 2b and 2c. = > Semi-Perimeter = ( 2a + 2b + 2c ) / 2 = > Semi-Perimeter = a + b + c By Using Heron's Formula : = > Area = √[ ( 2a + 2b + 2c )( 2a + 2b + 2c - 2a )( 2a + 2b + 2c - 2b )( 2a + 2b + 2c - 2c ) ] = > Area = √{ ( 2a + 2b + 2c )( 2b + 2c )( 2a + 2c )( 2a + 2b ) } = > Area = √{ 2( a + b + c )2( b + c )2( a + c )2( a + b ) } = > Area = 4√{ ( a + b + c )( b + c )( a + c )( a + b ) } When sides are doubled : Sides are 4a , 4b and 4c. = > Semi-Perimeter = ( 4a + 4b + 4c ) / 2 = > Semi-Perimeter = 2a + 2b + 2c By Using Heron's Formula : = > Area = √[ ( 4a + 4b + 4c )( 4a + 4b + 4c - 4a )( 4a + 4b + 4c - 4b )( 4a + 4b + 4c - 4c ) ] = > Area = √{ ( 4a + 4b + 4c )( 4b + 4c )( 4a + 4c )( 4a + 4b ) } = > Area = √{ 4( a + b + c )4( b + c )4( a + c )4( a + b ) } = > Area = 16√{ ( a + b + c )( b + c )( a + c )( a + b ) } Thus, Increase : 16√{ ( a + b + c )( b + c )( a + c )( a + b ) } - 4√{ ( a + b + c )( b + c )( a + c )( a + b ) } Increase : 12√{ ( a + b + c )( b + c )( a + c )( a + b ) } Increase % : [ 12√{ ( a + b + c )( b + c )( a + c )( a + b ) } ] / [ 4√{ ( a + b + c )( b + c )( a + c )( a + b ) } ] x 100% Increase % : 3 x 100% Increase % : 300% Hence the area is increased by 300 %.

45.

If each side of a triangle is doubled, find the % increase in the area of the triangle

Answer»

Answer:

Area is INCREASED by 300%.

Step-by-step explanation:

If that triangle is a right angled triangle.

= > Original area = 1 / 2 x base x height { From properties }

When sides are DOUBLED :

= > NEW area = 1 / 2 x 2base x 2height

= > New area = 2 x base x height

Increase % : ( 2 x height x base ) / ( 1 / 2 x base x height ) x 100%

= > 2 / ( 1 / 2 ) x 100%

= > 4 x 100%

= > 400%

Hence the area is increased by 400%.

If that triangle is not right angled triangle :

Let the sides are 2a , 2b and 2c.

= > Semi-Perimeter = ( 2a + 2b + 2c ) / 2

= > Semi-Perimeter = a + b + c

By Using Heron's Formula :

= > Area = √[ ( 2a + 2b + 2c )( 2a + 2b + 2c - 2a )( 2a + 2b + 2c - 2b )( 2a + 2b + 2c - 2c ) ]

= > Area = √{ ( 2a + 2b + 2c )( 2b + 2c )( 2a + 2c )( 2a + 2b ) }

= > Area = √{ 2( a + b + c )2( b + c )2( a + c )2( a + b ) }

= > Area = 4√{ ( a + b + c )( b + c )( a + c )( a + b ) }

When sides are doubled : Sides are 4a , 4b and 4c.

= > Semi-Perimeter = ( 4a + 4b + 4c ) / 2

= > Semi-Perimeter = 2a + 2b + 2c

By Using Heron's Formula :

= > Area = √[ ( 4a + 4b + 4c )( 4a + 4b + 4c - 4a )( 4a + 4b + 4c - 4b )( 4a + 4b + 4c - 4c ) ]

= > Area = √{ ( 4a + 4b + 4c )( 4b + 4c )( 4a + 4c )( 4a + 4b ) }

= > Area = √{ 4( a + b + c )4( b + c )4( a + c )4( a + b ) }

= > Area = 16√{ ( a + b + c )( b + c )( a + c )( a + b ) }

Thus,

Increase : 16√{ ( a + b + c )( b + c )( a + c )( a + b ) } - 4√{ ( a + b + c )( b + c )( a + c )( a + b ) }

Increase : 12√{ ( a + b + c )( b + c )( a + c )( a + b ) }

Increase % : [ 12√{ ( a + b + c )( b + c )( a + c )( a + b ) } ] / [ 4√{ ( a + b + c )( b + c )( a + c )( a + b ) } ] x 100%

Increase % : 3 x 100%

Increase % : 300%

Hence the area is increased by 300 %.

Discussion

46.	Write smallest 6-digit numberdifferent digits and haver unterten's place
Answer» Answer: 1,2,3,-1,-2-3 Step-by-step EXPLANATION: THINK this MAY be

46.

Write smallest 6-digit numberdifferent digits and haver unterten's place

Answer»

Answer:

1,2,3,-1,-2-3

Step-by-step EXPLANATION:

THINK this MAY be

Discussion

47.	Find the smallest 6 digit number which is exactly divisible by 210
Answer» 999810 is the RIGHT ANSWER of UR QUESTION

Discussion

48.	(iv) Jupiteris ninety eight lakh seventy five thousand kilometer from the siand Venus is twelve lakh elohi thousand kilometer from the Sun. Howmuch farther from the Sun is Venus?Solution
Answer» 98 75 000 + 12 11 000 11086000 is farther from the SUN by venus

48.

(iv) Jupiteris ninety eight lakh seventy five thousand kilometer from the siand Venus is twelve lakh elohi thousand kilometer from the Sun. Howmuch farther from the Sun is Venus?Solution

Answer»

98 75 000

+ 12 11 000

11086000 is farther from the SUN by venus

Discussion

49.	.Plot the points (-1,0) (1,0)(1,1)(0,2) and(-1,1).Join them inorder. What figure do you get?
Answer» Step-by-step EXPLANATION: we have a GOT a pentagon. u can ALSO CHECK by ploting on the GRAPH

49.

.Plot the points (-1,0) (1,0)(1,1)(0,2) and(-1,1).Join them inorder. What figure do you get?

Answer»

Step-by-step EXPLANATION:

we have a GOT a pentagon.

u can ALSO CHECK by ploting on the GRAPH

Discussion

50.	Define the term mole fraction
Answer» Step-by-step explanation: MOLE FRACTION..... ◆It is the amount of constituent ,DIVIDED by the amount of all the constituents. ■Here,there is a CONDITION, the SUM of all mole FRACTIONS is 1. ★The constituents are taken in mixture,n

50.

Define the term mole fraction

Answer»

Step-by-step explanation:

MOLE FRACTION.....

◆It is the amount of constituent ,DIVIDED by the amount of all the constituents.

■Here,there is a CONDITION, the SUM of all mole FRACTIONS is 1.

★The constituents are taken in mixture,n

Discussion

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