2 years ago

Solve v^3 = -26 where v is a real number. Simplify your answer as much as possible.

Mathematics

1 answer:

il63 [147K]2 years ago

5 0

Answer:

nearby -3

Step-by-step explanation:

When v^3 is -26, your v is equal to $-\sqrt[3]{26}$ .

3^3 is 27, so v is nearby 3.

or, It would be nearby 3.9

You might be interested in

Select the system of linear inequalities whose solution is graphed. <br><br> Please help !!!!

GrogVix [38]

$\boxed{y\geq 3x-2, \ x+2y\leq 4}$

<h2>Explanation:</h2>

First of all, we need to know the equations of the given lines. According to the options the equations of the lines are:

1. First equation: Written in Slope-Intercept Form:

$y=3x-2$

2. Second equation: Written in Standard Form:

$x+2y=4$

According to the graph, both lines are dashed meaning both inequalities must include either ≥ or ≤. So let's taste a point on the coordinate system in order to know which option is satisfied:

Hence, let's take point (0, 0):

$y \ (?) \ 3x-2 \\ \\ 0 \ (?) \ 3(0)-2 \\ \\ 0 \ (?) \ -2 \\ \\ \\ So \ 0 \ is \ greater \ than \ -2: \\ \\ 0\geq 2 \\ \\ \\ \\ x+2y \ (?) \ 4 \\ \\ 0+2(0) \ (?) \ 4 \\ \\ 0 \ (?) \ 4 \\ \\ \\ So \ 0 \ is \ less \ than \ 4: \\ \\ 0\leq 4$