332354 + Interview Questions in Secondary School in Math

1.	1. If 3 sin A + 5 cos A=5 then show that 5 sin A - 3 cos A= +3.(please provide full solution.)
Answer» ong>Answer: Answer ⤵️ Given 3 SIN A + 5 cos A = 5 Let 5 sin A - 3 cos A = x Squaring on both sides for both the equation 9 sin² A + 25 cos² A + 30 sin A cos A = 25 25 sin² A + 9 cos² A - 30 sin A cos A = x² Adding the equation 34( sin² A + cos² A ) = 25 + x² x² = 34-25 = 9 x= 3 5 sin A - 3 cos A 3 I hope answer is helpful to us is your RIGHT answer

1.

1. If 3 sin A + 5 cos A=5 then show that 5 sin A - 3 cos A= +3.(please provide full solution.)

Answer»

ong>Answer:

Answer ⤵️

Given 3 SIN A + 5 cos A = 5

Let 5 sin A - 3 cos A = x

Squaring on both sides for both the equation

9 sin² A + 25 cos² A + 30 sin A cos A = 25

25 sin² A + 9 cos² A - 30 sin A cos A = x²

Adding the equation

34( sin² A + cos² A ) = 25 + x²

x² = 34-25 = 9

x= 3

5 sin A - 3 cos A 3

I hope answer is helpful to us

is your RIGHT answer

Discussion

2.	The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.
Answer» align="absmiddle" alt="\HUGE{\bf{\green{\mathfrak{\MALTESE{Question:-}}}}}" class="latex-formula" ID="TexFormula1" src="https://tex.z-dn.net/?f=%20%5Chuge%7B%5Cbf%7B%5Cgreen%7B%5Cmathfrak%7B%5Cmaltese%7BQuestion%3A-%7D%7D%7D%7D%7D%20" title="\huge{\bf{\green{\mathfrak{\maltese{Question:-}}}}}"> $\bullet{\mapsto}$ The sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540cm. Find its area. $\huge {\bf{\orange{\mathfrak{\maltese{Answer:-}}}}}$ $\bullet{\leadsto} \: \textsf{Sides of triangle are in the ratio 12:17:25.}$ $\bullet{\leadsto} \: \textsf{Perimeter of the triangle = 540 cm.}$ $\bullet{\leadsto} \: \textsf{Let the sides of the triangle be x.}$ $\bullet{\leadsto} \: \therefore{\textsf{1st side = a = 12x.}}$ $\bullet{\leadsto} \: \therefore{\textsf{2nd side = b = 17x.}}$ $\bullet{\leadsto} \: \therefore{\textsf{3rd side = c = 25x.}}$ $\bullet{\leadsto} \: \textsf{Perimeter of the triangle = 540 cm.}$ $\bullet{\leadsto} \: \textsf{We know that,}$ $\bullet{\leadsto} \: \boxed{\textsf{Perimeter of a triangle = a + b + c.}}$ $\bullet{\leadsto} \: \textsf{Substituting the values,}$ $\bullet{\leadsto} \: \sf 540 \: = \: 12x \: + \: 17x \: + \: 25x$ $\bullet{\leadsto} \: \sf 540 \: = \: 54x$ $\bullet{\leadsto} \: \sf 54x \: = \: 540$ $\bullet{\leadsto} \: \sf x \: = \: \dfrac{540}{54}$ $\bullet{\leadsto} \: \boxed{\sf x \: = \: 10 \: cm.}$ $\bullet{\leadsto} \: \sf 1st \: side \: = \: a \: = \: 12x \: = \: 12 \: * \: 10 \: = \: 120 \: cm.$ $\bullet{\leadsto} \: \sf 2nd \: side \: = \: b \: = \: 17x \: = \: 17 \: * \: 10 \: = \: 170 \: cm.$ $\bullet{\leadsto} \: \sf 3rd \: side \: = \: c \: = \: 25x \: = \: 25 \: * \: 10 \: = \: 250 \: cm.$ $\bullet{\leadsto} \: \textsf{Using Heron's Formula,}$ $\bullet{\leadsto} \: \sf Semi-Perimeter \: = \: ?$ $\bullet{\leadsto} \: \sf Perimeter \: = \: 540 \: cm.$ $\bullet{\leadsto} \: \boxed{\sf Semi-Perimeter \: = \: \dfrac{Perimeter}{2}}$ $\bullet{\leadsto} \: \sf S \: = \: \dfrac{540}{2}$ $\bullet{\leadsto} \: \sf S \: = \: 270 \: cm.$ $\bullet{\leadsto} \: \boxed{\sf Heron's \: Formula \: = \: \sqrt{s(s \: - \: a)(s \: - \: b)(s \: - \: c)}.}$ $\bullet{\leadsto} \: \textsf{Substituting the values,}$ $\bullet{\leadsto} \: \sf \sqrt{270(270 \: - \: 120)(270 \: - \: 170)(270 \: - \: 250)}$ $\bullet{\leadsto} \: \sf \sqrt{270(150)(100)(20)}$ $\bullet{\leadsto} \: \sf \sqrt{270 \: * \: 150 \: * \: 100 \: * \: 20}$ $\bullet{\leadsto} \: \sf \sqrt{27 \: * \: 15 \: * \: 1 \: * \: 2 \: * \: 10^{5}}$ $\bullet{\leadsto} \: \sf \sqrt{810 \: * \: 10^{5}}$ $\bullet{\leadsto} \: \sf \sqrt{81 \: * \: 10^{6}}$ $\bullet{\leadsto} \: \sf 9 \: * \: 10^{3}$ $\bullet{\leadsto} \: \sf 9 \: * \: 1000$ $\bullet{\leadsto} \: \boxed{\sf 9,000 \: cm^{2}.}$ $\huge{\bf{\red{\mathfrak{\maltese{Conclusion:-}}}}}$ $\bullet{\mapsto} \: \boxed{\therefore{\sf Area \: of \: the \: triangle \: = \: 9,000 \: cm^{2}.}}$ $\huge{\bf{\purple{\mathfrak{\maltese{Formulas \: Used:-}}}}}$ $\bullet{\leadsto} \: \sf Heron's \: Formula \: = \: \sqrt{s(s \: - \: a)(s \: - \: b)(s \: - \: c)}.$ $\bullet{\leadsto} \: \sf Semi-Perimeter \: = \: \dfrac{Perimeter}{2} \: or \: \dfrac{a \: + \: b \: + \: c}{2}$ $\huge{\bf{\pink{\mathfrak{\maltese{Note:-}}}}}$ $\bullet{\leadsto}$ Here, to find the area of the triangle we don't know the height (h) of the triangle but we can USE Heron's Formula to find out the area of the triangle as we know the Semi-Perimeter (Half of the given perimeter) and the measure of all 3 sides.

2.

The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.

Answer»

align="absmiddle" alt="\HUGE{\bf{\green{\mathfrak{\MALTESE{Question:-}}}}}" class="latex-formula" ID="TexFormula1" src="https://tex.z-dn.net/?f=%20%5Chuge%7B%5Cbf%7B%5Cgreen%7B%5Cmathfrak%7B%5Cmaltese%7BQuestion%3A-%7D%7D%7D%7D%7D%20" title="\huge{\bf{\green{\mathfrak{\maltese{Question:-}}}}}">

$\bullet{\mapsto}$ The sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540cm. Find its area.

$\huge {\bf{\orange{\mathfrak{\maltese{Answer:-}}}}}$

$\bullet{\leadsto} \: \textsf{Sides of triangle are in the ratio 12:17:25.}$

$\bullet{\leadsto} \: \textsf{Perimeter of the triangle = 540 cm.}$

$\bullet{\leadsto} \: \textsf{Let the sides of the triangle be x.}$

$\bullet{\leadsto} \: \therefore{\textsf{1st side = a = 12x.}}$

$\bullet{\leadsto} \: \therefore{\textsf{2nd side = b = 17x.}}$

$\bullet{\leadsto} \: \therefore{\textsf{3rd side = c = 25x.}}$

$\bullet{\leadsto} \: \textsf{Perimeter of the triangle = 540 cm.}$

$\bullet{\leadsto} \: \textsf{We know that,}$

$\bullet{\leadsto} \: \boxed{\textsf{Perimeter of a triangle = a + b + c.}}$

$\bullet{\leadsto} \: \textsf{Substituting the values,}$

$\bullet{\leadsto} \: \sf 540 \: = \: 12x \: + \: 17x \: + \: 25x$

$\bullet{\leadsto} \: \sf 540 \: = \: 54x$

$\bullet{\leadsto} \: \sf 54x \: = \: 540$

$\bullet{\leadsto} \: \sf x \: = \: \dfrac{540}{54}$

$\bullet{\leadsto} \: \boxed{\sf x \: = \: 10 \: cm.}$

$\bullet{\leadsto} \: \sf 1st \: side \: = \: a \: = \: 12x \: = \: 12 \: * \: 10 \: = \: 120 \: cm.$

$\bullet{\leadsto} \: \sf 2nd \: side \: = \: b \: = \: 17x \: = \: 17 \: * \: 10 \: = \: 170 \: cm.$

$\bullet{\leadsto} \: \sf 3rd \: side \: = \: c \: = \: 25x \: = \: 25 \: * \: 10 \: = \: 250 \: cm.$

$\bullet{\leadsto} \: \textsf{Using Heron's Formula,}$

$\bullet{\leadsto} \: \sf Semi-Perimeter \: = \: ?$

$\bullet{\leadsto} \: \sf Perimeter \: = \: 540 \: cm.$

$\bullet{\leadsto} \: \boxed{\sf Semi-Perimeter \: = \: \dfrac{Perimeter}{2}}$

$\bullet{\leadsto} \: \sf S \: = \: \dfrac{540}{2}$

$\bullet{\leadsto} \: \sf S \: = \: 270 \: cm.$

$\bullet{\leadsto} \: \boxed{\sf Heron's \: Formula \: = \: \sqrt{s(s \: - \: a)(s \: - \: b)(s \: - \: c)}.}$

$\bullet{\leadsto} \: \textsf{Substituting the values,}$

$\bullet{\leadsto} \: \sf \sqrt{270(270 \: - \: 120)(270 \: - \: 170)(270 \: - \: 250)}$

$\bullet{\leadsto} \: \sf \sqrt{270(150)(100)(20)}$

$\bullet{\leadsto} \: \sf \sqrt{270 \: * \: 150 \: * \: 100 \: * \: 20}$

$\bullet{\leadsto} \: \sf \sqrt{27 \: * \: 15 \: * \: 1 \: * \: 2 \: * \: 10^{5}}$

$\bullet{\leadsto} \: \sf \sqrt{810 \: * \: 10^{5}}$

$\bullet{\leadsto} \: \sf \sqrt{81 \: * \: 10^{6}}$

$\bullet{\leadsto} \: \sf 9 \: * \: 10^{3}$

$\bullet{\leadsto} \: \sf 9 \: * \: 1000$

$\bullet{\leadsto} \: \boxed{\sf 9,000 \: cm^{2}.}$

$\huge{\bf{\red{\mathfrak{\maltese{Conclusion:-}}}}}$

$\bullet{\mapsto} \: \boxed{\therefore{\sf Area \: of \: the \: triangle \: = \: 9,000 \: cm^{2}.}}$

$\huge{\bf{\purple{\mathfrak{\maltese{Formulas \: Used:-}}}}}$

$\bullet{\leadsto} \: \sf Heron's \: Formula \: = \: \sqrt{s(s \: - \: a)(s \: - \: b)(s \: - \: c)}.$

$\bullet{\leadsto} \: \sf Semi-Perimeter \: = \: \dfrac{Perimeter}{2} \: or \: \dfrac{a \: + \: b \: + \: c}{2}$

$\huge{\bf{\pink{\mathfrak{\maltese{Note:-}}}}}$

$\bullet{\leadsto}$ Here, to find the area of the triangle we don't know the height (h) of the triangle but we can USE Heron's Formula to find out the area of the triangle as we know the Semi-Perimeter (Half of the given perimeter) and the measure of all 3 sides.

Discussion

3.	Rohan made 10/3 kg of biscuits. if he packs 2/12 kg in one packet, how many packets will be require to pack the whole amount
Answer» ONG>Step-by-step EXPLANATION: HELLO here is your ANSWER

Discussion

4.	43. 20 men can do a work in 39 days. In how many days 26 men will do the same work?(a) 25 days (b) 30 days (c)35days (d) 20 days
Answer» ONG>Answer: 30DAYS Step-by-step explanation: 20×39=26 = 30days

4.

43. 20 men can do a work in 39 days. In how many days 26 men will do the same work?(a) 25 days (b) 30 days (c)35days (d) 20 days

Answer» ONG>Answer:

30DAYS

Step-by-step explanation:

20×39=26

= 30days

Discussion

5.	5 अंको की सबसे छोटी संख्या निकालिए जिसमें 579 से भाग देने पर 124 शेष बचे
Answer» ONG>ANSWER: I don't have Hindi Step-by-step EXPLANATION: ok byee

5.

5 अंको की सबसे छोटी संख्या निकालिए जिसमें 579 से भाग देने पर 124 शेष बचे

Answer» ONG>ANSWER:

I don't have Hindi

Step-by-step EXPLANATION:

ok byee

Discussion

6.	5) Evaluare:Sin2 30 cos2 45° + 4 tan2 30
Answer» ONG>Answer: 37/24 Step-by-step EXPLANATION: I HOPE it's HELPFUL

6.

5) Evaluare:Sin2 30 cos2 45° + 4 tan2 30

Answer» ONG>Answer:

37/24

Step-by-step EXPLANATION:

I HOPE it's HELPFUL

Discussion

7.	Is (x-3)^2=-2(x+6) a quadratic equation
Answer»

Discussion

8.	If 28 Kg 700 g tomatoes are packed equally in 7 baskets, then how much tomatoes are there in each basket?
Answer» ong>Answer: Total TOMATOES :- 28 kg 700 g No. of BASKETS :- 7 $\blue{\bold{\underline{\underline{}}}}$ We need to divide weight of tomatoes by no, of baskets :- $\begin{gathered} :\implies\sf \dfrac { 28 kg \: 700 g \: } { 7 } \\ \\ \\ :\implies\sf\boxed{ 4 kg \:100 g \:}\end{gathered}$ Convert it in grams :- 1 kg = 1000 GM : $\implies$ 4000 + 100 grams : $\implies$ 4100 grams !

8.

If 28 Kg 700 g tomatoes are packed equally in 7 baskets, then how much tomatoes are there in each basket?

Answer»

ong>Answer:

Total TOMATOES :- 28 kg 700 g

No. of BASKETS :- 7

$\blue{\bold{\underline{\underline{}}}}$

We need to divide weight of tomatoes by no, of baskets :-

$\begin{gathered} :\implies\sf \dfrac { 28 kg \: 700 g \: } { 7 } \\ \\ \\ :\implies\sf\boxed{ 4 kg \:100 g \:}\end{gathered}$

Convert it in grams :- 1 kg = 1000 GM

: $\implies$ 4000 + 100 grams

: $\implies$ 4100 grams !

Discussion

9.	Find side and perimeter of square whose diagonal is 13√2
Answer» ONG>Answer: 17 Step-by-step explanation: 67/46 򐄨.46 25

9.

Find side and perimeter of square whose diagonal is 13√2

Answer» ONG>Answer:

17

Step-by-step explanation:

67/46

򐄨.46

25

Discussion

10.	find the determinant
Answer» ONG>ANSWER: EASY easy eagjnvzchbcsfbcafhGh

Discussion

11.	Two roads of width 3m run perpendicular to each other in the middle of a garden .The garden is of length 100m. The roads are laid at a cost of Rs 11 per m square and the remaining garden is levelled at Rs 5 per m square.Find the total amount spend
Answer» ONG>ANSWER: the answer will be Step-by-step EXPLANATION: RS 53546

11.

Two roads of width 3m run perpendicular to each other in the middle of a garden .The garden is of length 100m. The roads are laid at a cost of Rs 11 per m square and the remaining garden is levelled at Rs 5 per m square.Find the total amount spend

Answer» ONG>ANSWER:

the answer will be

Step-by-step EXPLANATION:

RS 53546

Discussion

12.	What is the RoundOFF7,09167
Answer» ONG>Answer: 709167 is ROUND of 709170

Discussion

13.	26. 2-50x 28. 4x2 - 9-30. (1 - x21-x2)(1 - y2) + 4xy32. 2(x - 2y) - 50y?34. X-136. 2x - 32
Answer» ong>Answer: I, §2, p. 13 1. - 3 < x < 3 2. - 1 ~ x ~ 0 3. - j3 ~ x ~ - 1 or 1 ~ x ~ j3 4. x < 3 or x > 7 5. - 1 < x < 2 6. x < -lor x > 1 7. - 5 < x < 5 8. - 1 ~ x ~ 0 9. x ~ 1 or x = 0 10. x ~ - 10 or x = 5 11. x ~ -10 or x = 5 12. x ~ 1 or x = -! 13. x < -4 14. -5 17. -2 < x < 8 18. 2 < x < 4 19. -4 < x < 10 20. x < -4 and x> 10 21. x < -10 and x > 4 I, §3, p. 17 1. -~ 2. (2x ~ 1) 3. 0, 2, 108 4. 2z - Z2, 2w - W2 5. x # j2 or -j2. f(5) = b 6. All x. f(27) = 3 7. (a) 1 (b) 1 (c) -1 (d) -1 8. (a) 1 (b) 4 (c) 0 (d) 0 9. (a) -2 (b) -6 (c) x2+4x-2 10. x~O, 2 II. (a) odd (b) even (c) odd (d) odd I, §4, p. 20 1. 8 and 9 2.! and -1 3. /6 and 2 4. ~ and 21/ 3 5. -h and! 6. 9 and 8 7. -! and -1 8.! and! 9. 1 and -! 10. -5:2

13.

26. 2-50x 28. 4x2 - 9-30. (1 - x21-x2)(1 - y2) + 4xy32. 2(x - 2y) - 50y?34. X-136. 2x - 32

Answer»

ong>Answer:

I, §2, p. 13

1. - 3 < x < 3 2. - 1 ~ x ~ 0 3. - j3 ~ x ~ - 1 or 1 ~ x ~ j3

4. x < 3 or x > 7 5. - 1 < x < 2 6. x < -lor x > 1 7. - 5 < x < 5

8. - 1 ~ x ~ 0 9. x ~ 1 or x = 0 10. x ~ - 10 or x = 5

11. x ~ -10 or x = 5 12. x ~ 1 or x = -! 13. x < -4

14. -5

17. -2 < x < 8 18. 2 < x < 4 19. -4 < x < 10 20. x < -4 and x> 10

21. x < -10 and x > 4

I, §3, p. 17

1. -~ 2. (2x ~ 1) 3. 0, 2, 108 4. 2z - Z2, 2w - W2

5. x # j2 or -j2. f(5) = b 6. All x. f(27) = 3

7. (a) 1 (b) 1 (c) -1 (d) -1 8. (a) 1 (b) 4 (c) 0 (d) 0

9. (a) -2 (b) -6 (c) x2+4x-2 10. x~O, 2

II. (a) odd (b) even (c) odd (d) odd

I, §4, p. 20

1. 8 and 9 2.! and -1 3. /6 and 2 4. ~ and 21/ 3 5. -h and! 6. 9 and 8

7. -! and -1 8.! and! 9. 1 and -! 10. -5:2

Discussion

14.	aman weighed 40 kg in the year 2005 , his weight increased by 10 % in the year 2006 and decreased 10% in the year 2007 what was his weight in 2007
Answer» ONG>Answer: 39.6 Step-by-step explanation: 40+4=44and-4.4=39.6

14.

aman weighed 40 kg in the year 2005 , his weight increased by 10 % in the year 2006 and decreased 10% in the year 2007 what was his weight in 2007

Answer» ONG>Answer:

39.6

Step-by-step explanation:

40+4=44and-4.4=39.6

Discussion

15.	Find the difference in the area of two squares, one with length of side 2.5m and the other with length of diagonal 5m. a:- 2.5sq.m b:-4sq.m c:- 6.35sq.m d:- 4.10sqm
Answer» A n s w e r ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ G i v e n ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Length of side of one square is 2.5 m Length of diagonal of other square is 5 m ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ F i n d ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ The difference in the area of TWO squares ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ S o l u t i o n ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ $\underline{ \underline{\bold{\texttt{Area of square :}}}}$ ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ ➠ Side × Side ⚊⚊⚊⚊ ⓵ ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ $\underline{ \underline{\bold{\texttt{Area of <klux>FIRST</klux> square :}}}}$ ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Side = 2.5 m ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ ⟮ Putting the VALUE in ⓵ ⟯ ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ Side × Side ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ 2.5 × 2.5 ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ 6.25 sq. m ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Hence the area of first square is 6.25 sq. m ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ $\underline{ \underline{\bold{\texttt{Diagonal of square :}}}}$ ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ ➠ Side² + Side² = Diagonal² ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Diagonal = 5 m ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ ⟮ Putting the above value in ⓶ ⟯ ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ Side² + Side² = Diagonal² ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ 2Side² = 5² ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ Side² = 25⁄2 ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ Side² = 12.5 ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Also , ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ Side² = Area of square ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Hence the area of second square is 12.5 sq. m ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ $\underline{ \underline{\bold{\texttt{Difference in the areas of two squares :}}}}$ ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Area of second square - Area of first square ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Area of second square = 12.5 sq. m Area of first square = 6.25 sq. m ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ ⟮ Putting the above values ⟯ ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Area of second square - Area of first square ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : ➜ 12.5 - 6.25 ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ : : ➨ 6.25 sq. m ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ Hence the difference in areas of two squares is 6.25 sq. m

15.

Find the difference in the area of two squares, one with length of side 2.5m and the other with length of diagonal 5m. a:- 2.5sq.m b:-4sq.m c:- 6.35sq.m d:- 4.10sqm

Answer»

A n s w e r

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G i v e n

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Length of side of one square is 2.5 m
Length of diagonal of other square is 5 m

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F i n d

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The difference in the area of TWO squares

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S o l u t i o n

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

$\underline{ \underline{\bold{\texttt{Area of square :}}}}$

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➠ Side × Side ⚊⚊⚊⚊ ⓵

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

$\underline{ \underline{\bold{\texttt{Area of <klux>FIRST</klux> square :}}}}$

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Side = 2.5 m

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⟮ Putting the VALUE in ⓵ ⟯

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: ➜ Side × Side

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: ➜ 2.5 × 2.5

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: ➜ 6.25 sq. m

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Hence the area of first square is 6.25 sq. m

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

$\underline{ \underline{\bold{\texttt{Diagonal of square :}}}}$

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

➠ Side² + Side² = Diagonal² ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Diagonal = 5 m

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

⟮ Putting the above value in ⓶ ⟯

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: ➜ Side² + Side² = Diagonal²

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: ➜ 2Side² = 5²

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: ➜ Side² = 25⁄2

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: ➜ Side² = 12.5

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Also ,

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: ➜ Side² = Area of square

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Hence the area of second square is 12.5 sq. m

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

$\underline{ \underline{\bold{\texttt{Difference in the areas of two squares :}}}}$

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Area of second square - Area of first square

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Area of second square = 12.5 sq. m
Area of first square = 6.25 sq. m

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

⟮ Putting the above values ⟯

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Area of second square - Area of first square

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: ➜ 12.5 - 6.25

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

: : ➨ 6.25 sq. m

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Hence the difference in areas of two squares is 6.25 sq. m

Discussion

16.	Solve by middle term splitting methodx²-4x-12
Answer» rong>ANSWER : x² - 4x - 12 = (X + 2) (x - 6) GIVEN : x² - 4x - 12 Solution : 》x² - 4x - 12 》x² - 6x + 2x - 12 》x(x - 6) + 2(x - 6) 》(x + 2) (x - 6) Hence , x² - 4x - 12 = (x + 2) (x - 6) or 》x² - 4x - 12 》x² - 6x + 2x - 12 》(x² - 6x) + (2x - 12) 》x(x - 6) + 2(x - 6) 》(x - 6) (x + 2)

16.

Solve by middle term splitting methodx²-4x-12

Answer»

rong>ANSWER :

x² - 4x - 12 = (X + 2) (x - 6)

GIVEN :

x² - 4x - 12

Solution :

》x² - 4x - 12

》x² - 6x + 2x - 12

》x(x - 6) + 2(x - 6)

》(x + 2) (x - 6)

Hence , x² - 4x - 12 = (x + 2) (x - 6)

or

》x² - 4x - 12

》x² - 6x + 2x - 12

》(x² - 6x) + (2x - 12)

》x(x - 6) + 2(x - 6)

》(x - 6) (x + 2)

Discussion

17.	9.By 40,592 how much exceeds 1,509
Answer» ONG>Answer:39083 Step-by-step EXPLANATION:

17.

9.By 40,592 how much exceeds 1,509

Answer» ONG>Answer:39083

Step-by-step EXPLANATION:

Discussion

18.	What is class mark between 35-45
Answer» ONG>Answer: 40 Step-by-step EXPLANATION: this is the answer hope you have understood 45+35÷2

18.

What is class mark between 35-45

Answer» ONG>Answer:

40

Step-by-step EXPLANATION:

this is the answer

hope you have understood

45+35÷2

Discussion

19.	3. Find the value of k, if x = 2, y = 1 is a solution of the equation2x + 3y = k.
Answer» ONG>ANSWER: I HOPE this will HELP you

19.

3. Find the value of k, if x = 2, y = 1 is a solution of the equation2x + 3y = k.

Answer» ONG>ANSWER:

I HOPE this will HELP you

Discussion

20.	B.190625 19062.5 x 100000= 7.9062 x 104it is correct plls answer I will mark as brain list
Answer» ong>Answer: PLEASE you CHECK your QUESTION .

20.

B.190625 19062.5 x 100000= 7.9062 x 104it is correct plls answer I will mark as brain list

Answer»

ong>Answer:

PLEASE you CHECK your QUESTION .

Discussion

21.	Angle Dab + angle abc + angle BCD + angle CDA
Answer» ONG>ANSWER: ufkfrlrkdkkfkfjfjfjfjfjfjfjrjfndndndnfnd Step-by-step explanation: join o. n. l. y. G. i. r. l. s for s. h. o. w. i. n. g b. o. b. s

21.

Angle Dab + angle abc + angle BCD + angle CDA

Answer» ONG>ANSWER:

ufkfrlrkdkkfkfjfjfjfjfjfjfjrjfndndndnfnd

Step-by-step explanation:

join o. n. l. y. G. i. r. l. s for s. h. o. w. i. n. g b. o. b. s

Discussion

22.	(3a2+4a-5)+(2a2 -3a+7) how to solve
Answer» TION:- 3a^2+2a^2+4a-3a-5+7 5a^2+a-2

22.

(3a2+4a-5)+(2a2 -3a+7) how to solve

Answer» TION:-

3a^2+2a^2+4a-3a-5+7

5a^2+a-2

Discussion

23.	The angle of elevation of anGeraplane fromthe groundis Goo After a flight of 30 seconds,the elevation changes toaeroplane is flying at a constantheight at 4500 metres. Find the speedof the plane
Answer» ONG>Answer: ANGLE 60 degree Step-by-step EXPLANATION: MAKE me as BRAINLIST

23.

The angle of elevation of anGeraplane fromthe groundis Goo After a flight of 30 seconds,the elevation changes toaeroplane is flying at a constantheight at 4500 metres. Find the speedof the plane

Answer» ONG>Answer:

ANGLE 60 degree

Step-by-step EXPLANATION:

MAKE me as BRAINLIST

Discussion

24.	3. 30 mm Make a concave side of a long arm using a semi-circular method.
Answer» ONG>ANSWER: anggyjgddtygfssgggggd

Discussion

25.	9. If Q is the set of rational numbers Ő is theset of irrational numbers, thena) QU Q=Rb) QcQc) Q CQ=Qd) None of these
Answer» ong>ANSWER: d is the ANSWERS in this QUESTIONS okk Step-by-step explanation: PLZ MARK me a brain lisr

25.

9. If Q is the set of rational numbers Ő is theset of irrational numbers, thena) QU Q=Rb) QcQc) Q CQ=Qd) None of these

Answer»

ong>ANSWER:

d is the ANSWERS in this QUESTIONS okk

Step-by-step explanation:

PLZ MARK me a brain lisr

Discussion

26.	please guys answer me I will mark you as brainlist answer please it's urgent don't give wrong answere its request
Answer» ONG>Answer: Will you be my FRIEND have a great day Step-by-step explanation: 6543210,1023456,3012456

26.

please guys answer me I will mark you as brainlist answer please it's urgent don't give wrong answere its request

Answer» ONG>Answer:

Will you be my FRIEND

have a great day

Step-by-step explanation:

6543210,1023456,3012456

Discussion

27.	(5root7 +3)(4-root7)
Answer» ong>Answer: 17√7 - 23 Step-by-step explanation: (5√7 + 3)(4 - √7) REMEMBER: 5√7 x 4 = 20√7 , 3 x √7 = 3√7 and √7 x √7 = 7 (5√7 + 3)(4 - √7) = 20√7 - 5(7) + 12 - 3√7 = 20√7 - 35 + 12 - 3√7 = 17√7 - 23

27.

(5root7 +3)(4-root7)

Answer»

ong>Answer:

17√7 - 23

Step-by-step explanation:

(5√7 + 3)(4 - √7)

REMEMBER:

5√7 x 4 = 20√7 ,

3 x √7 = 3√7

and √7 x √7 = 7

(5√7 + 3)(4 - √7)

= 20√7 - 5(7) + 12 - 3√7

= 20√7 - 35 + 12 - 3√7

= 17√7 - 23

Discussion

28.	COS (A + B). 5. (i) If sin A =VIOsin B =115107prove that A + B = "; A and B being positive acute4angles.
Answer» ong>Answer: answer is pie by four please mark my answer as brainliest. Step-by-step explanation: sin A = 1/√10 and sin B = 1/√5 We KNOW that SIN2X + cos2x = 1 Thus, cos A = √(1 – sin2A) = 3/√10 cos B = √(1 – SIN2B) = 2/√5 As we know, sin(A + B) = sin A cos B + cos A sin B = (1/√10) (2/√5) + (3/√10) (1/√5) = (2 + 3)/ √50 = 5/(5√2) = 1/√2 sin(A + B) = sin π/4 {since A and B are acute angles as per the given} Therefore, A + B = π/4

28.

COS (A + B). 5. (i) If sin A =VIOsin B =115107prove that A + B = "; A and B being positive acute4angles.

Answer»

ong>Answer:

answer is pie by four please mark my answer as brainliest.

Step-by-step explanation:

sin A = 1/√10 and sin B = 1/√5

We KNOW that SIN2X + cos2x = 1

Thus, cos A = √(1 – sin2A) = 3/√10

cos B = √(1 – SIN2B) = 2/√5

As we know,

sin(A + B) = sin A cos B + cos A sin B

= (1/√10) (2/√5) + (3/√10) (1/√5)

= (2 + 3)/ √50

= 5/(5√2)

= 1/√2

sin(A + B) = sin π/4 {since A and B are acute angles as per the given}

Therefore, A + B = π/4

Discussion

29.	What should be added to y2 + 4x − 9 to get 3x + 5
Answer» ONG>ANSWER: The QUESTION is not PROPER

Discussion

30.	The sum is from chp ratio and proportion. Please send a photo of the sum when done. Thank you
Answer» ong>ANSWER: ye loge SABKE answer delete KARTE HAI

30.

The sum is from chp ratio and proportion. Please send a photo of the sum when done. Thank you

Answer»

ong>ANSWER:

ye loge SABKE answer delete KARTE HAI

Discussion

31.	Solve √3÷3. pls explain
Answer» ONG>ANSWER: The answer is 0.57735026919 Step-by-step EXPLANATION: √3÷3 =>√1 =>0.57735026919

31.

Solve √3÷3. pls explain

Answer» ONG>ANSWER:

The answer is 0.57735026919

Step-by-step EXPLANATION:

√3÷3

=>√1

=>0.57735026919

Discussion

32.	X^2-x+3/(x-2)(x^2+1)
Answer» ONG>Step-by-step EXPLANATION: ∆ABC~ ∆DEF, if ⦟A = 45° and ⦟E = 35° then ⦟B =? (A) 45° (B)35° (C)25° (D) 40°

32.

X^2-x+3/(x-2)(x^2+1)

Answer» ONG>Step-by-step EXPLANATION:

∆ABC~ ∆DEF, if ⦟A = 45° and ⦟E = 35° then ⦟B =?

(A) 45° (B)35° (C)25° (D) 40°

Discussion

33.	If 5 students sit in a row, then the probability that two particular students are always together is
Answer» ONG>ANSWER: jdnf DJ rjdnsjbeje j rkd id dineiehe dkneod

Discussion

34.	It rained for 18 days last Januarywhat percentage of the month wasit dry?
Answer» ONG>ANSWER: (13/31) * 100 Step-by-step EXPLANATION:

34.

It rained for 18 days last Januarywhat percentage of the month wasit dry?

Answer» ONG>ANSWER:

(13/31) * 100

Step-by-step EXPLANATION:

Discussion

35.	How many angles are there in letter j
Answer» ONG>ANSWER: there are 2 angles in J letter

Discussion

36.	Write factors using prime factorization method : 215
Answer» ong>Answer: 215 and Level 3 215 is a composite number. Prime FACTORIZATION: 215 = 5 x 43. The exponents in the prime factorization are 1 and 1. ... Factors of 215: 1, 5, 43, 215. Factor PAIRS: 215 = 1 x 215 or 5 x 43. Since 215 has no SQUARE factors, the square root of 215 cannot be simplified.

36.

Write factors using prime factorization method : 215

Answer»

ong>Answer:

215 and Level 3

215 is a composite number.

Prime FACTORIZATION: 215 = 5 x 43.

The exponents in the prime factorization are 1 and 1. ...

Factors of 215: 1, 5, 43, 215.

Factor PAIRS: 215 = 1 x 215 or 5 x 43.

Since 215 has no SQUARE factors, the square root of 215 cannot be simplified.

Discussion

37.	Construct aa square maudiagonal Sumaoparea.andfind its
Answer» ong>Answer: bhai SAHI SE questions likho pahele question SAMAJ me NAHI aa smraha

37.

Construct aa square maudiagonal Sumaoparea.andfind its

Answer»

ong>Answer:

bhai SAHI SE questions likho pahele question SAMAJ me NAHI aa smraha

Discussion

38.	(h) If a number exactly divides two numbers separately, it must exactly divide their sum.(1) Ifa number exactly divides the sum of two numbers, it must exactly divide the two
Answer» ONG>Answer: Really I don't KNOW SORRY buddy

Discussion

39.	The total surface area of a solid hemisphere of radius r is ------ a) 2pi r^22 b) 3pi r^22 c) 4pi r^22 d) pi r^22
Answer» ong>ANSWER: b) 3pi r^2 Step-by-step EXPLANATION: The total surface AREA of a SOLID HEMISPHERE of radius r is ------ 3 pi r^2 Hope this will help you

39.

The total surface area of a solid hemisphere of radius r is ------ a) 2pi r^22 b) 3pi r^22 c) 4pi r^22 d) pi r^22

Answer»

ong>ANSWER:

b) 3pi r^2

Step-by-step EXPLANATION:

The total surface AREA of a SOLID HEMISPHERE of radius r is ------

3 pi r^2

Hope this will help you

Discussion

40.	Help me solving these questions. plz plz don't scam. those who don't know the answer please don't answer
Answer» ong>Answer: xccdffctxexexu the the the the the the is a good time to explore NEW ways of LEARNING and SMART ways of learning and smart ways of learning and smart

40.

Help me solving these questions. plz plz don't scam. those who don't know the answer please don't answer

Answer»

ong>Answer:

xccdffctxexexu the the the the the the is a good time to explore NEW ways of LEARNING and SMART ways of learning and smart ways of learning and smart

Discussion

41.	the length of two perpendicular sides of a right triangle are 3cm and 2cm a)calculate the length of the hypothesis. b)lf a square is drawn, with this hypothesis as it's side then. 1)area of the square is ________cm.sq. 2)perimeter of the square is_________cm
Answer» ONG>Step-by-step EXPLANATION: swq-ragv-ukm no bo.obs required only gi.rls.. 17 CM

41.

the length of two perpendicular sides of a right triangle are 3cm and 2cm a)calculate the length of the hypothesis. b)lf a square is drawn, with this hypothesis as it's side then. 1)area of the square is cm.sq. 2)perimeter of the square is_cm

Answer» ONG>Step-by-step EXPLANATION:

swq-ragv-ukm

no bo.obs required

only gi.rls..

17 CM

Discussion

42.	If sin0 + cos0 = p and sin³0 + cos³0 = q , then find the value of 3p - 2q. Please answer it.
Answer» ONG>Step-by-step EXPLANATION: NNSA and its MEMBERS were 9inches

Discussion

43.	1. A refrigerator was bought for 12000 and sold at 13000. Find theprofit and profit percent.
Answer» ONG>Answer: 1. A refrigerator was bought for 12000 and SOLD at 13000. Find the PROFIT and profit percent.

43.

1. A refrigerator was bought for 12000 and sold at 13000. Find theprofit and profit percent.

Answer» ONG>Answer:

1. A refrigerator was bought for 12000 and SOLD at 13000. Find the

PROFIT and profit percent.

Discussion

44.	Form an a quadratic equation from x = .... where x is a natural number.
Answer» ONG>Answer: please write✍️✍️✍️ the FULL QUESTION

Discussion

45.	In figure ABCD is a cyclic quadrilateral if DAB=75° then find the measure DCB
Answer» ong>Answer: 105 Step-by-step EXPLANATION: OPPOSITE ANGLES are supplementary of cyclic quadrilateral. ANGLE DCB+ angle DAB=180° angle DCB+75°=180° angle DCB=180°-75° angle DCB=105°

45.

In figure ABCD is a cyclic quadrilateral if DAB=75° then find the measure DCB

Answer»

ong>Answer:

105

Step-by-step EXPLANATION:

OPPOSITE ANGLES are supplementary of cyclic quadrilateral.

ANGLE DCB+ angle DAB=180°

angle DCB+75°=180°

angle DCB=180°-75°

angle DCB=105°

Discussion

46.	Chinmay invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get after 1 and half year?
Answer» align="absmiddle" alt="{\pmb{\underline{\sf{ Required \ Solution ... }}}} \\" CLASS="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%20%7B%5Cpmb%7B%5Cunderline%7B%5Csf%7B%20Required%20%5C%20Solution%20...%20%7D%7D%7D%7D%20%5C%5C%20" title="{\pmb{\underline{\sf{ Required \ Solution ... }}}} \\"> Principal = ₹ 60,000 Rate (r) = 12% (Compounded half yearly) Time (t) = 1 & ½ YEARS (3 parts) We have to adjust Rate and time According to the half yearly compounded as:- $\circ \ {\pmb{\underline{\boxed{\sf{ Amount = P \left( 1 + \dfrac{r}{100} \right)^ n }}}}} \\$ $\\ \bigstar \ {\pmb{\underline{\sf{ According \ to \ Question: }}}} \\ \\ \\ \colon\implies{\sf{ 60000 \left( 1 + \dfrac{6}{100} \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \left( \cancel{ \dfrac{106}{100} } \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \left( \dfrac{53}{50} \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \times \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50} }} \\ \\ \\ \colon\implies{\sf{ 60 \cancel{000} \times \dfrac{53}{5 \cancel{0} } \times \dfrac{53}{5 \cancel{0} } \times \dfrac{53}{5 \cancel{0} } }} \\ \\ \\ \colon\implies{\sf{ 60 \times \dfrac{53}{5} \times \dfrac{53}{5} \times \dfrac{53}{5} }} \\ \\ \\ \colon\implies{\sf{ \cancel{ \dfrac{8932620}{125} } }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf\green{ Rs. \ 71460.96 _{(Amount)} }}}} \\$ HENCE, $\\ {\pmb{\underline{\sf\green{ The \ Amount \ of \ the \ Investment \ is \ Rs. \ 71460.96. }}}}$

46.

Chinmay invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get after 1 and half year?

Answer»

align="absmiddle" alt="{\pmb{\underline{\sf{ Required \ Solution ... }}}} \\" CLASS="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%20%7B%5Cpmb%7B%5Cunderline%7B%5Csf%7B%20Required%20%5C%20Solution%20...%20%7D%7D%7D%7D%20%5C%5C%20" title="{\pmb{\underline{\sf{ Required \ Solution ... }}}} \\">

Principal = ₹ 60,000
Rate (r) = 12% (Compounded half yearly)
Time (t) = 1 & ½ YEARS (3 parts)

We have to adjust Rate and time According to the half yearly compounded as:-

$\circ \ {\pmb{\underline{\boxed{\sf{ Amount = P \left( 1 + \dfrac{r}{100} \right)^ n }}}}} \\$

$\\ \bigstar \ {\pmb{\underline{\sf{ According \ to \ Question: }}}} \\ \\ \\ \colon\implies{\sf{ 60000 \left( 1 + \dfrac{6}{100} \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \left( \cancel{ \dfrac{106}{100} } \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \left( \dfrac{53}{50} \right)^ 3 }} \\ \\ \\ \colon\implies{\sf{ 60000 \times \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50} }} \\ \\ \\ \colon\implies{\sf{ 60 \cancel{000} \times \dfrac{53}{5 \cancel{0} } \times \dfrac{53}{5 \cancel{0} } \times \dfrac{53}{5 \cancel{0} } }} \\ \\ \\ \colon\implies{\sf{ 60 \times \dfrac{53}{5} \times \dfrac{53}{5} \times \dfrac{53}{5} }} \\ \\ \\ \colon\implies{\sf{ \cancel{ \dfrac{8932620}{125} } }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf\green{ Rs. \ 71460.96 _{(Amount)} }}}} \\$

HENCE,

$\\ {\pmb{\underline{\sf\green{ The \ Amount \ of \ the \ Investment \ is \ Rs. \ 71460.96. }}}}$

Discussion

47.	30% of 120 fruits are rotten find the number of fruits rotten
Answer» ONG>Answer: PLZ MARK me as BRAINLEST Step-by-step explanation: $30 \div 100 \times 120$ then the answer will come that will the no. of rotten fruits

47.

30% of 120 fruits are rotten find the number of fruits rotten

Answer» ONG>Answer:

PLZ MARK me as BRAINLEST

Step-by-step explanation:

$30 \div 100 \times 120$

then the answer will come that will the no. of rotten fruits

Discussion

48.	10) If A=4x2 + y2 + 7xy, B=7x2 – 6y2 – 10xy, C=x2 + y2 + xyfind i) A-B+C ii) 2A+C.
Answer» ong>ANSWER: *Hope it helps!! MARK this answer as BRAINLIEST if u found it useful and FOLLOW* *me for quick and ACCURATE answers...* Step-by-step explanation: $\bold\red{A=4x {}^{2} + y {}^{2} + 7xy} \\ \\ \bold\blue{B=7x {}^{2} - 6y {}^{2} - 10xy }\\ \\ \bold\pink{C=x{}^{2} + y {}^{2} + xy} \\ - - - - - - - - - - \\ i) A-B+C \\ \\ \red{4x {}^{2} + y {}^{2} + 7xy} - (\blue{7x {}^{2} - 6y {}^{2} - 10xy } ) + \pink{x{}^{2} + y {}^{2} + xy} \\ \\ \red{4x {}^{2} + y {}^{2} + 7xy} \blue{ - 7x {}^{2} + 6y {}^{2} + 10xy } \pink{ + x{}^{2} + y {}^{2} + xy} \\ \\ 4x {}^{2} - 7 {x}^{2} + {x}^{2} + y {}^{2} + 6y {}^{2} + {y}^{2} + 7xy + 10xy + xy \\ \\ \bold\green{ - 2 {x}^{2} + 8 {y}^{2} + 18xy} \\ - - - - - - - - - - \\ ii) 2A+C\\ \\ 2(\red{4x {}^{2} + y {}^{2} + 7xy}) + \pink{x{}^{2} + y {}^{2} + xy} \\ \\ \red{8 {x}^{2} + 2 {y}^{2} + 14xy} + \pink{x{}^{2} + y {}^{2} + xy} \\ \\ 8 {x}^{2} + {x}^{2} + 2 {y}^{2} + {y}^{2} + 14xy + xy \\ \\\bold\green{ 9 {x}^{2} + 3 {y}^{2} + 15 xy}$

48.

10) If A=4x2 + y2 + 7xy, B=7x2 – 6y2 – 10xy, C=x2 + y2 + xyfind i) A-B+C ii) 2A+C.

Answer»

ong>ANSWER:

Hope it helps!! MARK this answer as BRAINLIEST if u found it useful and FOLLOW me for quick and ACCURATE answers...

Step-by-step explanation:

$\bold\red{A=4x {}^{2} + y {}^{2} + 7xy} \\ \\ \bold\blue{B=7x {}^{2} - 6y {}^{2} - 10xy }\\ \\ \bold\pink{C=x{}^{2} + y {}^{2} + xy} \\ - - - - - - - - - - \\ i) A-B+C \\ \\ \red{4x {}^{2} + y {}^{2} + 7xy} - (\blue{7x {}^{2} - 6y {}^{2} - 10xy } ) + \pink{x{}^{2} + y {}^{2} + xy} \\ \\ \red{4x {}^{2} + y {}^{2} + 7xy} \blue{ - 7x {}^{2} + 6y {}^{2} + 10xy } \pink{ + x{}^{2} + y {}^{2} + xy} \\ \\ 4x {}^{2} - 7 {x}^{2} + {x}^{2} + y {}^{2} + 6y {}^{2} + {y}^{2} + 7xy + 10xy + xy \\ \\ \bold\green{ - 2 {x}^{2} + 8 {y}^{2} + 18xy} \\ - - - - - - - - - - \\ ii) 2A+C\\ \\ 2(\red{4x {}^{2} + y {}^{2} + 7xy}) + \pink{x{}^{2} + y {}^{2} + xy} \\ \\ \red{8 {x}^{2} + 2 {y}^{2} + 14xy} + \pink{x{}^{2} + y {}^{2} + xy} \\ \\ 8 {x}^{2} + {x}^{2} + 2 {y}^{2} + {y}^{2} + 14xy + xy \\ \\\bold\green{ 9 {x}^{2} + 3 {y}^{2} + 15 xy}$

Discussion

49.	12 मीटर ऊंचे एक खंबे की जमीन पर छाया की लंबाई 9 मीटर है।उसी समय पर 18 मीटर ऊंचे खंबे की छाया कीलंबाई होगी :- The length of the shadow of apole12 meters high is 9 meters on the ground. Atthe same time the length of the shadow of thepole 18 meters high will be: -
Answer» ER: STEP by step explanation refer to IMAGE ATTACHED

49.

12 मीटर ऊंचे एक खंबे की जमीन पर छाया की लंबाई 9 मीटर है।उसी समय पर 18 मीटर ऊंचे खंबे की छाया कीलंबाई होगी :- The length of the shadow of apole12 meters high is 9 meters on the ground. Atthe same time the length of the shadow of thepole 18 meters high will be: -

Answer» ER:

STEP by step explanation

refer to IMAGE ATTACHED

Discussion

50.	Find the median of the following data: 7, 10, 7, 5, 9, 10 ?
Answer» ONG>Step-by-step EXPLANATION: $7 + 10 + 7 + 5 + 9 + 10 = 48$ $48 \div 6 = 8$

50.

Find the median of the following data: 7, 10, 7, 5, 9, 10 ?

Answer» ONG>Step-by-step EXPLANATION:

$7 + 10 + 7 + 5 + 9 + 10 = 48$

$48 \div 6 = 8$

Discussion

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REMEMBER:

5√7 x 4 = 20√7 ,

3 x √7 = 3√7

and √7 x √7 = 7

(5√7 + 3)(4 - √7)

= 20√7 - 5(7) + 12 - 3√7

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= 17√7 - 23

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