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IgorC [24]
3 years ago
12

I can't figure out the attached problem

Mathematics
1 answer:
MrMuchimi3 years ago
7 0
The y goes down 5 every time x goes up 1. The equation would be y=-5x-1
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Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 4 h, and Car B traveled the dista
poizon [28]
The rate of Car A is x
The rate of Car B is x - 5

They travel the same distance. So we can make an equation:

4(x) = 4.5(x-5)

Solve for x.

4(x) = 4.5(x-5)
4x = 4.5x - 22.5

Subtract 4.5x on both sides

-0.5x = -22.5

Divide by -0.5 on both sides
x = 45

But we want the speed of Car B which is x - 5

So Car B's speed is 45 - 5 = 40

Your answer is 40
8 0
3 years ago
If the complex number x = 3 + bi and |x|2 = 13, which is a possible value of b?
Naddika [18.5K]

Answer:

A: 2

Step-by-step explanation:

EDGE 2021

3 0
2 years ago
Which expression is equivalent to...? Screenshots attached, please help.
BaLLatris [955]

For this case we must simplify the following expression:

\sqrt [3] {64 * a ^ 6 * b ^ 7 * c ^ 9}

We rewrite:

64 = 4 ^ 3\\a ^ 6 = (a ^ 2) ^ 3\\b ^ 6 * b = ((b ^ 2) ^ 3 * b)\\c ^ 9 = (c ^ 3) ^ 3

So:

\sqrt [3] {4 ^ 3 * (a ^ 2) ^ 3 * (b ^ 2) ^ 3 * b * (c ^ 3) ^ 3} =\\\sqrt [3] {4 * a ^ 2 * b ^ 2 * c ^ 3) ^ 3 * b} =

By definition of properties of powers and roots we have:

\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}

So:

4a ^ 2b ^ 2 c ^ 3 \sqrt [3] {b}

Answer:

Option B

3 0
3 years ago
Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions
quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of v_0 and F_2 be the component normal to v_0.

Since, |v_0| = 1,

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, v_0 = \left

From figure,

|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3

We know that the direction of F_1 is opposite of the direction of v_0, so we have

$F_1 = -5\sqrt3 v_0$

    $=-5\sqrt3 \left$

    $= \left$

The unit vector in the direction normal to the plane, v_1 has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5

∴  F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>

                   $=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

           $= = F$

3 0
2 years ago
(7x^8-10v)-(4x^8+2v)
olchik [2.2K]
3 ( x^8-4v) ........... .
6 0
3 years ago
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