What is 4 less than 5 by the 3 power equal

Please help I dont understand

Kitty [74]

Answer:

3rd option

Step-by-step explanation:

A direct variation is expressed in this form:

y = kx

1. y - 4x = 8

y = 4x + 8 is not a direct variation

2. y + 2 = 7x

y = 7x - 2 is not a direct variation

3. y - 3x = 0

y = 3x is a direct variation

4. y = 5x - 2

y = 5x - 2 is not direct variation

8 0

2 years ago

Simplify (-2x^4y^-3)^4

elixir [45]

Answer:

$( - 2 {x}^{4} {y}^{ - 3} )^{4} = \frac{16 {x}^{16}}{ {y}^{12} }$

Step-by-step explanation:

We want to simplify:

$( - 2 {x}^{4} {y}^{ - 3} )^{4}$

We need to apply the exponential property for products of powers.

Recall that:

$( {a}^{m} \times {a}^{n} )^{p} = {a}^{mp} \times {a}^{np}$

We apply this rule to get:

$( - 2 {x}^{4} {y}^{ - 3} )^{4} = ( { - 2)}^{4} x \times {x}^{16} \times {y}^{ - 12}$

This simplifies to:

$( - 2 {x}^{4} {y}^{ - 3} )^{4} = 16 {x}^{16} \times {y}^{ - 12}$

We rewrite as positive index to get:

$( - 2 {x}^{4} {y}^{ - 3} )^{4} = \frac{16 {x}^{16}}{ {y}^{12} }$

7 0

3 years ago

Which statements are true about the interquartile range? Select all that apply.

kap26 [50]

Is there a picture? there's not enough info

4 0

2 years ago

Read 2 more answers

A party rental company has chairs and tables for rent. The total cost to rent 5 chairs and 3 tables is $31.the total cost to ren

san4es73 [151]

Answer:

$1.25 & $10.25

Step-by-step explanation:

Let c be the cost of renting one chair and t be the cost of renting table. We're given two equations:

#1. 5c + 3t = 37

#2. 2c + 6t =64

We have a system of equations. Using our system of equations calculator, we can solve this problem any of 3 ways below:

Chairs (c) cost $1.25
Tables (t) cost $10.25

6 0

2 years ago

Can someone please help me on number 16-ABC

melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

$-2x\:$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)>10\left(-1\right)$

Simplify

$2x>-10$

Divide both sides by 2

$\frac{2x}{2}>\frac{-10}{2}$

$x>-5$

$-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}$

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

$-2x>-6$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)$

Simplify

$2x$

Divide both sides by 2

$\frac{2x}{2}$

$x$

$-6$

Thus, the two intervals:

$\left(-\infty \:,\:3\right)$

$\left(-5,\:\infty \:\right)$

The intersection of these two intervals would be the solution to both inequalities.

$\left(-\infty \:,\:3\right)$ and $\left(-5,\:\infty \:\right)$

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0

2 years ago