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4 years ago

Simplify the expression where possible. (x^2)^4

Mathematics

1 answer:

balu736 [363]4 years ago

3 0

Answer:

Step-by-step explanation:

According to the third law of indices,

In this case m=2 and n=4

So that can be applied thus:

2×4=8

so it will now be written this way:

x^8

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If you can answer these TRICK, yes I said TRICK questions, I will give you brainliest. Yes, brainliest.

Fynjy0 [20]

Answer:

what are they? I need the brainliest ;-;

8 0

3 years ago

Read 2 more answers

One number is seven less than a second number. Five times the first is 2 more than 6 times the second. Find the numbers. The val

creativ13 [48]

Answer:

$\Large \boxed{\sf \ \ 40 \ \ }$

Step-by-step explanation:

Hello,

Let's not a and b the two numbers.

One number is seven less than a second number.

(1) b = a - 7

Five times the first is 2 more than 6 times the second.

(2) 5a = 2 + 6b

(2)+5*(1) gives

5a + 5b = 2 + 6b + 5a - 35

b = 35 - 2 = 33

a = 40

The value of the first number is $\large \boxed{\sf \ \ 40 \ \ }$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

4 0

3 years ago

Divide the fractions: 2 / 5 ÷ 3 / 4.

Serjik [45]

Answer:

8/15

Step-by-step explanation:

KEEP FLIP FLIP

keep 2/5

flip the sign (multiply, not divide)

flip 3/4 to 4/3

2/5 x 4/3

8/15

4 0

3 years ago

How many subsets are there in {1, 2, 3, 4, 5, 6}?

Contact [7]

<h3>Answer: 64</h3>

Explanation:

If there are n = 6 items in the set, then there are 2^n = 2^6 = 64 subsets

This is because there are 2 choices per slot. Either the item is in the subset or not. Since we have 6 slots, and 2 choices per slot, we have (2*2*2)*(2*2*2) = 2^6 = 64 different combos.

Side note: The empty set is included as one of the subsets. Also, the original set itself is a subset.

8 0

3 years ago

Read 2 more answers

What are all the real zeros

xeze [42]

For this case we have a function of the form $y = f (x)$ , where $f (x) = (x-12) ^ 3-10$

To find the real zeros we must equal zero and clear the variable "x".

$(x-12) ^ 3-10 = 0$

We add 10 to both sides of the equation

$(x-12) ^ 3-10 + 10 = 10\\(x-12) ^ 3 = 10$

We apply cube root to both sides of the equation:

$\sqrt[3]{(x-12)^3} = \sqrt[3] {10}\\x-12 = \sqrt[3] {10}$

We add 12 to both sides of the equation:

$x-12 + 12 = \sqrt[3] {10} +12\\x = \sqrt[3] {10} +12$

Answer:

Option D

5 0

3 years ago