332354 + Interview Questions in Secondary School in Math

1.	Town of 10000 families it is found that 40% families by newspaper a 20% families by newspaper be 10% families by newspaper 65% families by newspaper A and B 3% families by newspaper B and C 4% families by newspaper a and c if two families boy all the three newspaper find the number of families which by newspaper A onlyOnly answer was needed
Answer» ANSWER: Step-by-step EXPLANATION: 3300 .... ....

1.

Town of 10000 families it is found that 40% families by newspaper a 20% families by newspaper be 10% families by newspaper 65% families by newspaper A and B 3% families by newspaper B and C 4% families by newspaper a and c if two families boy all the three newspaper find the number of families which by newspaper A onlyOnly answer was needed

Answer»

ANSWER:

Step-by-step EXPLANATION:

3300

....

Discussion

2.	Please answerPlease please please please please
Answer» WELCOME to the CONCEPT of IMPLICIT function. I hope it help you OTHERWISE REPORT it...

2.

Please answerPlease please please please please

Answer»

WELCOME to the CONCEPT of IMPLICIT function.

I hope it help you OTHERWISE REPORT it...

Discussion

3.	Find the product of 0and -64
Answer» ANSWER: Any number multiplied by 0 gives the result 0. Therefore, 0-64=0 0 is the* *RIGHT* *answer* *Hope* it *HELPS* ✌✌

3.

Find the product of 0and -64

Answer»

ANSWER:

Any number multiplied by 0 gives the result 0.

Therefore,

0*-64=0

0 is the RIGHT answer

Hope it HELPS ✌✌

Discussion

4.	Please answer the given questions :1. In Triangle ABC AD is perpendicular to BC. (AD)2= BD X CD. prove that AB2+AC2=(BD +CD) 2.
Answer» Answer: In the GIVEN Fig., AD ⊥ BC. PROVE that AB2 + CD2 = BD2 + AC2. In right ∆ADC. by PYTHAGORAS Theorem, AC2 = AD2 + CD2 ...(i) In right ∆ADB, by Pythagoras Theorem, AB2 = AD2 + BD2 ...(ii) Subtracting (i) from (ii), we get AB2 - AC2 = BD2 - CD2 ⇒ AB2 + CD2 = BD2 + AC2 Hence PROVED.

4.

Please answer the given questions :1. In Triangle ABC AD is perpendicular to BC. (AD)2= BD X CD. prove that AB2+AC2=(BD +CD) 2.

Answer»

Answer:

In the GIVEN Fig., AD ⊥ BC. PROVE that AB2 + CD2 = BD2 + AC2.

In right ∆ADC. by PYTHAGORAS Theorem,

AC2 = AD2 + CD2 ...(i)

In right ∆ADB, by Pythagoras Theorem,

AB2 = AD2 + BD2 ...(ii)

Subtracting (i) from (ii), we get

AB2 - AC2 = BD2 - CD2

⇒ AB2 + CD2 = BD2 + AC2 Hence PROVED.

Discussion

5.	Which subject or course suitable after 10th to take the course BCA after 12th
Answer» ANSWER: COMMERCE with BIOLOGY MEDICAL

5.

Which subject or course suitable after 10th to take the course BCA after 12th

Answer»

ANSWER:

COMMERCE with BIOLOGY MEDICAL

Discussion

6.	What is the meaning of 69 apart from number??
Answer» 69 is slang for when TWO partners arrange their BODIES to perform oral SEX on ONE another at the same time in a way said to look like the number 69.......

Discussion

7.	amar collected 2157 marbles and karan collected 9293 marblaes how many marbles do they own collected?
Answer» AMAN COLLECTED = 2157 Karan collected = 9293 they own collected = 11450

7.

amar collected 2157 marbles and karan collected 9293 marblaes how many marbles do they own collected?

Answer»

AMAN COLLECTED = 2157

Karan collected = 9293

they own collected = 11450

Discussion

8.	What should be added to 25/3to get46/5
Answer» 46/5-25/3 ((463)-(255))/15 (138-75)/15 63/15= ....21/5 should be ADDED

8.

What should be added to 25/3to get46/5

Answer»

46/5-25/3

((46*3)-(25*5))/15

(138-75)/15

63/15=

....21/5 should be ADDED

Discussion

9.	If a+b=11and ab=28 then find the value of a cube + b cube...
Answer» Answer: hey here is your answer mate. Step-by-step EXPLANATION: let a=4 b=7 then,a+b=11. ab=4×7=28 a cube + b cube =(a+b)(a SQUARE -ab+b square) 43+73=(4+7)(42-28+72) =11×(16-28+49) =11×(-12+49) =11×37 =407(answer) ANSWER IS COMPLETED.

9.

If a+b=11and ab=28 then find the value of a cube + b cube...

Answer»

Answer:

hey here is your answer mate.

Step-by-step EXPLANATION:

let a=4 b=7

then,a+b=11. ab=4×7=28

a cube + b cube =(a+b)(a SQUARE -ab+b square)

43+73=(4+7)(42-28+72)

=11×(16-28+49)

=11×(-12+49)

=11×37

=407(answer)

ANSWER IS COMPLETED.

Discussion

10.	(-9)-(-6)=? help me please
Answer» Step-by-step EXPLANATION: (-9)-(-6) As (-)(-)=+ -9-(-6)=-9+6 -3 is the* *RIGHT* *answer* *HOPE* it *HELPS* ✌✌

10.

(-9)-(-6)=? help me please

Answer»

Step-by-step EXPLANATION:

(-9)-(-6)

As (-)*(-)=+

-9-(-6)=-9+6

-3 is the RIGHT answer

HOPE it HELPS ✌✌

Discussion

11.	Plzzz answer this question ❤❤❤ Guys I need your urgent help !! ❤❤☺
Answer» ANSWER: PUT x=a in equation a= -2 because it is a factor of eqn then x-a must be EQUAL to 0

11.

Plzzz answer this question ❤❤❤ Guys I need your urgent help !! ❤❤☺

Answer»

ANSWER:

PUT x=a in equation

a= -2

because it is a factor of eqn then x-a must be EQUAL to 0

Discussion

12.	A number is doubled and 9 is added. if the result is trebled it becomes 75. what is that number.
Answer» Answer: 75/3 =25 25-9=16 16/2 = 8 MAKE it BRAINLIST PLZ

12.

A number is doubled and 9 is added. if the result is trebled it becomes 75. what is that number.

Answer»

Answer:

75/3 =25

25-9=16

16/2 = 8 MAKE it BRAINLIST PLZ

Discussion

13.	Add bracket 3 root 2 + 7 root 3 and root 2 minice root 3
Answer» ANSWER: (3√2 + 7√3) + ( √2 - 5√3) = 3√2 + 7√3 + √2 - 5√3 = 3√2 + √2 + 7√3 - 5√3 = 4√2 + 2√3

13.

Add bracket 3 root 2 + 7 root 3 and root 2 minice root 3

Answer»

ANSWER:

(3√2 + 7√3) + ( √2 - 5√3)

= 3√2 + 7√3 + √2 - 5√3

= 3√2 + √2 + 7√3 - 5√3

= 4√2 + 2√3

Discussion

14.	5 men can do a peice of work in 6 dayswhile 10 women can do it in 5 days. inhow many days can 5 women and 3 mendo it.
Answer» ANSWER: it TAKES 6.25 DAYS.......

14.

5 men can do a peice of work in 6 dayswhile 10 women can do it in 5 days. inhow many days can 5 women and 3 mendo it.

Answer»

ANSWER:

it TAKES 6.25 DAYS.......

Discussion

15.	Pls answer step by step.....
Answer» Step-by-step EXPLANATION: i not understamd the PICTURE

15.

Pls answer step by step.....

Answer»

Step-by-step EXPLANATION:

i not understamd the PICTURE

Discussion

16.	What is algebra?tell some examples
Answer» HEY MATE ANSWER THE MATHEMATICAL TOPIC IN WHICH THE LETTERS AND SYMBOLS ARE USED TO DINOTE NUMBERS ARE CALLED ALGEBRA PLEASE FOLLOW ME PLEASE ITS A REQUEST

16.

What is algebra?tell some examples

Answer»

HEY MATE

ANSWER

THE MATHEMATICAL TOPIC IN WHICH THE LETTERS AND SYMBOLS ARE USED TO DINOTE NUMBERS ARE CALLED ALGEBRA

PLEASE FOLLOW ME PLEASE ITS A REQUEST

Discussion

17.	The cost of 5 kg oranges is Rs. 450. The cost of 12dozen Banana is Rs. 4320 and the cost of 4 kgpears is Rs. 240. What is the cost of 8 kg oranges,8 dozen Banana and 8 kg pears?
Answer» Cost of 5kg oranges is RS 450 cost of 1kg oranges is 450/5= RS 90 ✓cost of 8kg oranges is 908= RS 720 cost of 12 DOZEN banana is RS 4320 cost of 1 dozen banana is 4320/12= RS 360 cost of 8 dozen banana is 3608 = RS 2880 cost of 4kg pears is RS 240 cost of 1kg PEAR is 240/4 = RS 60 cost of 8kg pear is 60*8 = RS 480

Discussion

18.	4. What is the multiplicative inverse of a negative rational number?
Answer» The RECIPROCAL of that particular negative RATIONAL number is the multiplicaticative INVERSE. For EG -9/5 is the multiplicative inverse of -5/9. Hope it helps!!! :)

18.

4. What is the multiplicative inverse of a negative rational number?

Answer»

The RECIPROCAL of that particular negative RATIONAL number is the multiplicaticative INVERSE. For EG -9/5 is the multiplicative inverse of -5/9.

Hope it helps!!! :)

Discussion

19.	X =2+root3 find value of x cube +1/ x cube
Answer» ANSWER : This is UR answer

19.

X =2+root3 find value of x cube +1/ x cube

Answer»

ANSWER : This is UR answer

Discussion

20.	Quadratic polynomial 2xsquare-3x+1 has zeros as alpha and beta now form quadratic equation whose roots are 3alpha and 3 beta
Answer» DIVYA sangwan a GENIUS GIRL

Discussion

21.	Which no used in algebra.
Answer» ANSWER: VARIABLE and CONSTANT both are USED in ALGEBRA.

21.

Which no used in algebra.

Answer»

ANSWER:

VARIABLE and CONSTANT both are USED in ALGEBRA.

Discussion

22.	Keeping potential difference constant the resistance is 1/3.Then by how much does the current changes.
Answer» Step-by-step EXPLANATION: CURRENT will be TRIPLE

22.

Keeping potential difference constant the resistance is 1/3.Then by how much does the current changes.

Answer»

Step-by-step EXPLANATION:

CURRENT will be TRIPLE

Discussion

23.	Find the mode and mean for the set of data below
Answer» PLEASE EDIT THE QUESTION BECAUSE WHERE ARE THE NUMBERS ?????

23.

Find the mode and mean for the set of data below

Answer»

PLEASE EDIT THE QUESTION BECAUSE WHERE ARE THE NUMBERS ?????

Discussion

24.	Find the area of an isosceles triangle with two equal sides as 7 cm each and the third side as 6 cm.
Answer» Step-by-step EXPLANATION: here is the ANSWER you NEEDED for this QUESTION

24.

Find the area of an isosceles triangle with two equal sides as 7 cm each and the third side as 6 cm.

Answer»

Step-by-step EXPLANATION:

here is the ANSWER you NEEDED for this QUESTION

Discussion

25.	Find the value of k so that the lines 2x-3y=9 and kx-9y=18 will be parallel
Answer» $\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$ TO FIND :- The VALUE of k. Its Given :- Lines are parallel. 2x - 3y = 9 kx - 9Y = 18 For Intersecting lines :- a₁/a₂ ≠ b₁/b₂. ( UNIQUE solution ) For coincident lines :- a₁/a₂ = b₁/b₂ = c₁/c₂. ( Infinite solution ) For Parallel lines :- a₁/a₂ = b₁/b₂ ≠ c₁/c₂. ( No solution ) 2x - 3y -9 = 0 kx - 9y -18 = 0 a₁ = 2 , b₁ = -3 , c₁ = -9 a₂ = k , b₂ = -9 , c₂ = -18 a₁/a₂ = b₁/b₂ ≠ c₁/c₂ 2/k = 3/9 ≠ 9/18 3k = 18 k = 18/3 k = 6. Hence, the value of "k" = 6.

25.

Find the value of k so that the lines 2x-3y=9 and kx-9y=18 will be parallel

Answer»

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

TO FIND :- The VALUE of k.

Its Given :- Lines are parallel.

2x - 3y = 9

kx - 9Y = 18

For Intersecting lines :- a₁/a₂ ≠ b₁/b₂. ( UNIQUE solution )

For coincident lines :- a₁/a₂ = b₁/b₂ = c₁/c₂. ( Infinite solution )

For Parallel lines :- a₁/a₂ = b₁/b₂ ≠ c₁/c₂. ( No solution )

2x - 3y -9 = 0

kx - 9y -18 = 0

a₁ = 2 , b₁ = -3 , c₁ = -9

a₂ = k , b₂ = -9 , c₂ = -18

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

2/k = 3/9 ≠ 9/18

3k = 18

k = 18/3

k = 6.

Hence, the value of "k" = 6.

Discussion

26.	720 divide in 5 person and person e get 40 rs more a distrubtion
Answer» I36,136,136,136,176.are the ANS

Discussion

27.	Find the value of [(525+252)(525252)]/ (525252). (a) 3 (b) 4 (c) 5 (d) 6
Answer» Step-by-step EXPLANATION: $\frac{(525 + 252) \times 525252}{525252} \\ 525 + 252 \\ = 357$

27.

Find the value of [(525+252)(525252)]/ (525252). (a) 3 (b) 4 (c) 5 (d) 6

Answer»

Step-by-step EXPLANATION:

$\frac{(525 + 252) \times 525252}{525252} \\ 525 + 252 \\ = 357$

Discussion

28.	Find the roots by the method of completing the square 2xsquare-7x+3
Answer» Step-by-step EXPLANATION: ${2x}^{2} - 7x + <klux>3</klux> \\ {2x}^{2} - 6x - 1x + 3 \\ 2x(x - 3) - 1(x - 3) \\( 2x - 1)(x - 3)$ ROOTS are 1/2 and 3

28.

Find the roots by the method of completing the square 2xsquare-7x+3

Answer»

Step-by-step EXPLANATION:

${2x}^{2} - 7x + <klux>3</klux> \\ {2x}^{2} - 6x - 1x + 3 \\ 2x(x - 3) - 1(x - 3) \\( 2x - 1)(x - 3)$

ROOTS are 1/2 and 3

Discussion

29.	Find minimum distance between circle and straight line
Answer» Step-by-step explanation: the minimum DISTANCE between the CIRCLE and straight line is zero please mark it as BRAIN LIST please

29.

Find minimum distance between circle and straight line

Answer»

Step-by-step explanation:

the minimum DISTANCE between the CIRCLE and straight line is zero

please mark it as BRAIN LIST please

Discussion

30.	Arrange them in ascending order of magnitude3√3,3√4,5√2
Answer» ANSWER: 3√3,3√4,5√2 it will HELP U

30.

Arrange them in ascending order of magnitude3√3,3√4,5√2

Answer»

ANSWER:

3√3,3√4,5√2

it will HELP U

Discussion

31.	Find the product of 16 and -12
Answer» 16*-12= .....-192..........

31.

Find the product of 16 and -12

Answer»

16*-12=

.....-192..........

Discussion

32.	5-23= R-113, find a and b.7-43
Answer» ANSWER: a=11 b= -6 Step-by-step EXPLANATION: LOOK the IMAGE

32.

5-23= R-113, find a and b.7-43

Answer»

ANSWER:

a=11

b= -6

Step-by-step EXPLANATION:

LOOK the IMAGE

Discussion

33.	Is sin theta = 1= cos theta possible?give reason.
Answer» Yes BCOZ TAN THETA =1 in case of 45 DEGREE

Discussion

34.	258÷25dome this with full process...
Answer» HEYA.. HOPE it HELPS you BUDDY.. FOLLOW me ✌

Discussion

35.	51,95,54,319 in international system in words
Answer» Answer: Five hundred NINETEEN million five hundred FIFTY four THOUSAND three hundred nineteen. I Hope This MAY HELP U.

35.

51,95,54,319 in international system in words

Answer»

Answer:

Five hundred NINETEEN million five hundred FIFTY four THOUSAND three hundred nineteen.

I Hope This MAY HELP U.

Discussion

36.	Vas SIMILAR TRIANGLES CBSE 2GR = 1+ 2IN APOR, ST 11 QR. PT=4 CM, TR=Fund. Ar (PST): ar (PQR).
Answer» Step-by-step EXPLANATION: i DONT UNDERSTAND your QUESTION

36.

Vas SIMILAR TRIANGLES CBSE 2GR = 1+ 2IN APOR, ST 11 QR. PT=4 CM, TR=Fund. Ar (PST): ar (PQR).

Answer»

Step-by-step EXPLANATION:

i DONT UNDERSTAND your QUESTION

Discussion

37.	If x^4 + X^3 +X^2 +1=0 then find X^2015 + X^3015 + 5=?
Answer» ANSWER: We GET x=1 then 1^2015+1^3015+5=7.

37.

If x^4 + X^3 +X^2 +1=0 then find X^2015 + X^3015 + 5=?

Answer»

ANSWER:

We GET x=1 then

1^2015+1^3015+5=7.

Discussion

38.	Show that the square of any position odd.integer is of the form 8 m +1 for someintegerm
Answer» ANSWER: HEY guys Step-by-step EXPLANATION: This TYPE MARK AS BRAINLIST the

38.

Show that the square of any position odd.integer is of the form 8 m +1 for someintegerm

Answer»

ANSWER:

HEY guys

Step-by-step EXPLANATION:

This TYPE

MARK AS BRAINLIST the

Discussion

39.	Oldest building of indus valley civilization
Answer» Answer: BUILT around 2600 BCE, it was one of the largest settlements of the ANCIENT Indus Valley Civilization, and one of the world's earliest major urban settlements, CONTEMPORANEOUS with the CIVILIZATIONS of ancient Egypt, Mesopotamia, and Crete.

39.

Oldest building of indus valley civilization

Answer»

Answer:

BUILT around 2600 BCE, it was one of the largest settlements of the ANCIENT Indus Valley Civilization, and one of the world's earliest major urban settlements, CONTEMPORANEOUS with the CIVILIZATIONS of ancient Egypt, Mesopotamia, and Crete.

Discussion

40.	9. किसी कार्य को A उतने समय में कर सकता है।/ जितने समय में B और C मिलकर उस कार्यको कर सकते हैं। यदि A और B इस कार्यको मिलकर 10 दिन में कर सकते हैं तथा Cअकेला उसे 50 दिन में कर सकता है, तो Bअकेला उस कार्य को कितने समय में करपाएगा?(1) 15 दिन (2) 20 दिन ।(3) 25 दिन (4) 30 दिन
Answer» Step-by-step explanation: i think 20 DIN mein KAR SAKTA HAI.

40.

9. किसी कार्य को A उतने समय में कर सकता है।/ जितने समय में B और C मिलकर उस कार्यको कर सकते हैं। यदि A और B इस कार्यको मिलकर 10 दिन में कर सकते हैं तथा Cअकेला उसे 50 दिन में कर सकता है, तो Bअकेला उस कार्य को कितने समय में करपाएगा?(1) 15 दिन (2) 20 दिन ।(3) 25 दिन (4) 30 दिन

Answer»

Step-by-step explanation:

i think 20 DIN mein KAR SAKTA HAI.

Discussion

41.	एक घनाकार डिब्बे का एक किनारा 10 cm लंबाई का है तथा एक अन्य घनाभाकार डिब्बे की लम्बाई, चौड़ाई और ऊँचाई क्रमशः 12.5 cm, 10 cm और 8 cm हैं।(i) किस डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल अधिक है और कितना अधिक है?(ii) किस डिब्बे का कुल पृष्ठीय क्षेत्रफल कम है और कितना कम है?
Answer» Answer: घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल 40 cm² अधिक है। घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल 10 cm² कम है। Step-by-step EXPLANATION: दिया है : एक घनाकार डिब्बे का एक किनारे की लंबाई (a) = 10 cm तथा एक अन्य घनाभाकार डिब्बे की लम्बाई (l) = 12.5 cm, चौड़ाई (b) = 10 cm और ऊँचाई (h) 8 cm (i) घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 4a² = 4 ×10² = 400 cm² घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 400 cm² घनाभाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 2(l + b) ×h = 2 × (12.5 + 10) × 8 = 2 × 22.5 × 8 = 360 cm² घनाभाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 360 cm² ∴ घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल अधिक है [∵ 400 > 360] अभीष्ट क्षेत्रफल = (400 – 360) = 40 cm² अतः ,घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल 40 cm² अधिक है (ii) घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 6a² = 6 × 10² = 600 cm² घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 600 cm² घनाभाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 2(lb + bh + lh) = 2(12.5 × 10 + 10×8 + 8 ×12.5) = 2(125 + 80 + 100) = (2 × 305) = 610 cm² घनाभाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 610 cm² ∴ घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल कम है। [∵ 610 > 600] अभीष्ट क्षेत्रफल = (610 – 600) = 10 cm² अतः , घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल 10 cm² कम है। आशा है कि यह उत्तर आपकी मदद करेगा। इस पाठ से संबंधित कुछ और प्रश्न : किसी आयताकार हॉल के फर्श का परिमाप 250 m है। यदि 10 प्रति $m2$ की दर से चारों दीवारों पर पेंट कराने की लागत रु 15000 है, तो इस हॉल की ऊँचाई ज्ञात कीजिए। [संकेत : चारों डिब्बों का क्षेत्रफल = पाश्र्व पृष्ठीय क्षेत्रफल ] brainly.in/question/10330762 किसी डिब्बे में भरा हुआ पेंट 9.375 $m^2$ के क्षेत्रफल पर पेंट करने के लिए पर्याप्त है। इस डिब्बे के पेट से 22.5 cm x 10 cm x7.5 cm विमाओं वाली कितनी ईंट पेंट की जा सकती हैं? brainly.in/question/10331218

41.

एक घनाकार डिब्बे का एक किनारा 10 cm लंबाई का है तथा एक अन्य घनाभाकार डिब्बे की लम्बाई, चौड़ाई और ऊँचाई क्रमशः 12.5 cm, 10 cm और 8 cm हैं।(i) किस डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल अधिक है और कितना अधिक है?(ii) किस डिब्बे का कुल पृष्ठीय क्षेत्रफल कम है और कितना कम है?

Answer»

Answer:

घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल 40 cm² अधिक है।

घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल 10 cm² कम है।

Step-by-step EXPLANATION:

दिया है :

एक घनाकार डिब्बे का एक किनारे की लंबाई (a) = 10 cm तथा एक अन्य घनाभाकार डिब्बे की लम्बाई (l) = 12.5 cm, चौड़ाई (b) = 10 cm और ऊँचाई (h) 8 cm

(i) घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 4a²

= 4 ×10²

= 400 cm²

घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 400 cm²

घनाभाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 2(l + b) ×h

= 2 × (12.5 + 10) × 8

= 2 × 22.5 × 8

= 360 cm²

घनाभाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 360 cm²

∴ घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल अधिक है [∵ 400 > 360]

अभीष्ट क्षेत्रफल = (400 – 360) = 40 cm²

अतः ,घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल 40 cm² अधिक है

(ii) घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 6a²

= 6 × 10²

= 600 cm²

घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 600 cm²

घनाभाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 2(lb + bh + lh)

= 2(12.5 × 10 + 10×8 + 8 ×12.5)

= 2(125 + 80 + 100)

= (2 × 305)

= 610 cm²

घनाभाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 610 cm²

∴ घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल कम है।

[∵ 610 > 600]

अभीष्ट क्षेत्रफल = (610 – 600) = 10 cm²

अतः , घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल 10 cm² कम है।

आशा है कि यह उत्तर आपकी मदद करेगा।

इस पाठ से संबंधित कुछ और प्रश्न :

किसी आयताकार हॉल के फर्श का परिमाप 250 m है। यदि 10 प्रति $m2$ की दर से चारों दीवारों पर पेंट कराने की लागत रु 15000 है, तो इस हॉल की ऊँचाई ज्ञात कीजिए।

[संकेत : चारों डिब्बों का क्षेत्रफल = पाश्र्व पृष्ठीय क्षेत्रफल ]

brainly.in/question/10330762

किसी डिब्बे में भरा हुआ पेंट 9.375 $m^2$ के क्षेत्रफल पर पेंट करने के लिए पर्याप्त है। इस डिब्बे के पेट से 22.5 cm x 10 cm x7.5 cm विमाओं वाली कितनी ईंट पेंट की जा सकती हैं?

brainly.in/question/10331218

Discussion

42.	Hieee guys !! Plzz help me !!The answer is 2 !! But I want step by step explanation !! I need your urgent help !! ❤❤❤
Answer» ANSWER: I think .....FIRSTLY ..do middle term SPLITTING of the equation so you will GET 2 FACTORS of that equation...then if any one factor is same as given above then choose the second one if it is different .....then divide the equation with that factor and you will get your remainder

42.

Hieee guys !! Plzz help me !!The answer is 2 !! But I want step by step explanation !! I need your urgent help !! ❤❤❤

Answer»

ANSWER:

I think .....FIRSTLY ..do middle term SPLITTING of the equation so you will GET 2 FACTORS of that equation...then if any one factor is same as given above then choose the second one if it is different .....then divide the equation with that factor and you will get your remainder

Discussion

43.	The population of a town increase every year by 2% of the population at the beginning of that year. The number of years by which the total increase of population be 40% is..
Answer» ANSWER: hii dear Step-by-step EXPLANATION: LET the base population=1000 + 2%increase=20% __________ 1020ans 1400=1000(1.02)^n 1.4=(1.02)^n n=17. PLZ MARK AS BRAINLIST

43.

The population of a town increase every year by 2% of the population at the beginning of that year. The number of years by which the total increase of population be 40% is..

Answer»

ANSWER:

hii dear

Step-by-step EXPLANATION:

LET the base population=1000

+ 2%increase=20%

__________

1020ans

1400=1000(1.02)^n

1.4=(1.02)^n

n=17.

PLZ MARK AS BRAINLIST

Discussion

44.	(vi)10 x (-12) + 5 X (-12)
Answer» (-120)+(-60) =-120-60 =-180

44.

(vi)10 x (-12) + 5 X (-12)

Answer»

(-120)+(-60)

=-120-60

=-180

Discussion

45.	A polynomial of degree four is called ----------•
Answer» A POLYNOMIAL of degree FOUR is called biquadratic or quartic EQUATIONS.

45.

A polynomial of degree four is called ----------•

Answer»

A POLYNOMIAL of degree FOUR is called biquadratic or quartic EQUATIONS.

Discussion

46.	23. The sum of the digits of two digit is 18. If the digits are reversed the number is 126 more the original. Find the number24. A Grandfather is ten times older than his granddaughter. He is also 54 years older than her. Findpresent ages.
Answer» Step-by-step EXPLANATION: Ans 2. LET the age of grandfather be = x The age of granddaughter be = y According to the question x = 10y x = 54 + y By elimination method -9y = - 54 y = 6years hence x= 60years

46.

23. The sum of the digits of two digit is 18. If the digits are reversed the number is 126 more the original. Find the number24. A Grandfather is ten times older than his granddaughter. He is also 54 years older than her. Findpresent ages.

Answer»

Step-by-step EXPLANATION:

Ans 2. LET the age of grandfather be = x

The age of granddaughter be = y

According to the question

x = 10y

x = 54 + y

By elimination method

-9y = - 54

y = 6years

hence x= 60years

Discussion

47.	the magnitude of vector A is 2 units and the vector B is 3 units.they are perpendicular to each other.the magnitude of 2A+B is?
Answer» Answer: 2a+b 2×2+3=7 UNITS...

47.

the magnitude of vector A is 2 units and the vector B is 3 units.they are perpendicular to each other.the magnitude of 2A+B is?

Answer»

Answer:

2a+b

2×2+3=7 UNITS...

Discussion

48.	Solve:5(z+4)=35 pls pls pls pls pls pls pls
Answer» ANSWER: z=3 is the answer because (5×3)+(5×4) 15+20 =35

48.

Solve:5(z+4)=35 pls pls pls pls pls pls pls

Answer»

ANSWER:

z=3

is the answer

because

(5×3)+(5×4)

15+20

=35

Discussion

49.	(-37)×(-7)+(-37)×(-3) which is the answer ,370,-370,148,,-148
Answer» *Heya❤ $( - 37) \times ( - 7) + ( - 37) \times ( - 3) \\ = 259 + 111 \\ = 370$ Hence; UR ANSWER is* 370 Hope it HELPS you ✌✌✌

49.

(-37)×(-7)+(-37)×(-3) which is the answer ,370,-370,148,,-148

Answer»

Heya❤

$( - 37) \times ( - 7) + ( - 37) \times ( - 3) \\ = 259 + 111 \\ = 370$

Hence;

UR ANSWER is 370

Hope it HELPS you

✌✌✌

Discussion

50.	draw the graph of x+y=5 and x-y=3on the same graph sheet.Find the coordinates of the point of intersection of the lines
Answer» ANSWER: Step-by-step EXPLANATION: OK it's DONE

50.

draw the graph of x+y=5 and x-y=3on the same graph sheet.Find the coordinates of the point of intersection of the lines

Answer»

ANSWER:

Step-by-step EXPLANATION:

OK it's DONE

Discussion

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घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल 40 cm² अधिक है।

घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल 10 cm² कम है।

घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 400 cm²

घनाभाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 360 cm²

अतः ,घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल 40 cm² अधिक है

घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 600 cm²

घनाभाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल = 610 cm²

अतः , घनाकार डिब्बे का संपूर्ण पृष्ठ क्षेत्रफल 10 cm² कम है।

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