All categories

3 years ago

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

Mathematics

1 answer:

balu736 [363]3 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

You might be interested in

Simplify the expression by combining like terms.

Tju [1.3M]

Answer:

8x + 13

Step-by-step explanation:

"Like terms" means the same kind of algebraic expression...the 6x and the 2x are "like terms" because they both have an x in them. The 9 and the 4 are both constants, which means they are plain numbers. So 6x and 2x is 8x .

9+4 is 13 just like elementary school.

So all together,

6x + 9 + 2x + 4

6x + 2x + 9 + 4

8x + 13

We can rearrange the terms because the problem has all addition in between the terms.

4 0

2 years ago

Solve 5/8+3/4 -2/3-5/6

alexira [117]

Answer:

-.125

Step-by-step explanation:

5/8 + 3/4 solve

2/3 - 5/6 solve

then subtract your numbers

3 0

3 years ago

Read 2 more answers

What is the distance from the point (3,8) to the point (7,-6)?

oksano4ka [1.4K]

Answer: Distance is 14.6

Step-by-step explanation:

$d = √((x2-x1)2 + (y2-y1)2)$

$(x2-x1) = (7 - 3) = 4$

$(y2-y1) = (-6 - 8) = -14$

add it all together

$(4)2 + (-14)2 = 16 + 196 = 212$

Square 212

$\sqrt{212} =14.5602$

round to the nearest tenth

=14.6

PLEASE MARK BRAINLIEST

5 0

3 years ago

Read 2 more answers

Note: You may use a calculator to solve the following problems.

Verizon [17]

Use a calculator it will help.

4 0

3 years ago

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th

irga5000 [103]

Answer:

The rocket hits the ground at a time of 11.59 seconds.

Step-by-step explanation:

The height of the rocket, after x seconds, is given by the following equation:

$y = -16x^2 + 177x + 98$

It hits the ground when $y = 0$ , so we have to find x for which $y = 0$ , which is a quadratic equation.

Finding the roots of a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$y = -16x^2 + 177x + 98$

$-16x^2 + 177x + 98 = 0$

$a = -16, b = 177, c = 98$

$\bigtriangleup = 177^{2} - 4(-16)(98) = 37601$

$x_{1} = \frac{-177 + \sqrt{37601}}{2*(-16)} = -0.53$

$x_{2} = \frac{-177 - \sqrt{37601}}{2*(-16)} = 11.59$

Since time is a positive measure, the rocket hits the ground at a time of 11.59 seconds.

4 0

3 years ago