Question

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Answer 1

Answer:

1] (47-25)+(52-47)×8

22+5×8=22+40=62

2] 1000÷[(150+50)÷10×(10-5)]

1000÷[200÷10×5]

1000÷[20×5]

1000÷100

10

3] 2^3÷(2+5-4)^2+5

8÷(3)^2+5

8÷9+5

0.88+5

5.88

4] (15-5)^3÷(8+2)^2+5

10^3÷10^2+5

10+5

15

5] 5^2÷(29-8÷4)^2×1^5

25÷(29-2)^2×1

25÷27^2

25÷729

0.0342

6] (5+4)^2÷(4+2÷2)^0+6^3

(20)^2÷4^0+216

400÷1+216

616

7] 100×{5^2+[(20-15+5)^2÷2]}÷75

100×{25+[10^2÷2]}÷75

100×{25+50}÷75

100

8] 3^3÷[(14-8+4)-1^3]

27÷[10-1]

27÷9

3

Answer 2

Step-by-step explanation:

ANSWER:-

1. (47-25)+(53-47)×8

For this, what we need to do is apply BODMAS.

What is BODMAS by the way?

$\boxed{\sf{BODMAS- What \ is \ this}}$ ?

It is actually Brackets Of Division Multiplication Addition Subtraction.

Now, let's solve off further.

We will solve the elements in the brackets first.

22+6×8

= 22+48

= 70 is the answer.

2. 1000÷[(150+50)÷10×(10-5)]

1000÷[(200)÷10×5]

= 1000÷[200÷50]

=1000÷[4]

= 25 is the answer.

3. 2³÷(2+5-4)²+5

= 8÷(7-4)²+5

= 8÷3²+5

= 8÷9+5

=0.88+5

= 5.88 (approximately, as its continuous like 5.8888 ..)

4. (15-5)³÷(8+2)²-5

= (10)³÷(10)²-5

$\sf{10^{(3-2)}-5}$

= 10 - 5

=5 is the answer.

5. 5²÷(29-8÷4)²×1⁵

1⁵ is nothing but 1 itself.

= 5²÷(29-2)²×1

= 25÷729

So, we can write the answer also as $\dfrac{25}{729}$

Or, $\sf{(\dfrac{5}{27})^2}$

6. (5+4)²-(4+2÷2)⁰+6³

Anything to the power zero is 1, so,

(9)² + 216 ____________[6³=216]

= 297 is the answer.

7. 100×{5²+[(20-15+5)²÷2]}÷75

= 100×{25 +(0)}÷75

=2500 ÷ 75

= 33.33 is the answer [Basically approx again, 33.33 bar]

8. 3³÷[(14-8+4)-1³]

= 27÷ [2-1]

= 27 is the required answer.