Choice 1:
= 5 * [5^3 / (2/5)]^2
= by looking at the equation, it's kinda not equal to 125

choice 2:
= [5^3 / 5^4]^-3
= 5^-9 / 5^-12
= 5^(-9 - -12)
= 5^3
= 125

choice 3:
= 5^-2 / 5^-5
= 5^(-2 - -5)
= 5^3
= 125

choice 4:
= 5 * 5^5/5^3
= 5 * 5^(5-3)
= 5 * 5^2
= 5^3
= 125

So the expression that is not equal to 125 is:
<span>5 times the quotient 5 cubed over two-fifths, raised to the second power</span>

5 0

3 years ago

The base of a triangle is 2 more than twice the height of the triangle. Find the measure of the base and the height if the area

vekshin1

B = 2 + 2H
A = 20 ft^2
H = H

Since the Area of a Triangle is Base * Height then the formula should be:
A = BH
20 = H(2+2H)
20 = 2H^2 + 2H
0 = 2H^2 + 2H - 20

Find the zeros of the quadratic equation through substituting the values of a,b, and c. You will get the zeros of x = 2.7 and -3.7. The only thing that would make sense is you use the x=2.7 because there are no negative heights.

Since b = 2h+2
b= 2(2.7) + 2
b= 5.4 + 2
b=7.4

I don't know where you got 4 and 10, but what I got is different.

8 0

3 years ago

If r(x) = 3x – 1 and s(x) = 2x + 1, which expression is equivalent to (r/s)<br>6)?

Flauer [41]

$\bf \begin{cases} r(x)=3x-1\\ s(x)=2x+1\\ \cline{1-1} \cfrac{r(x)}{g(x)}=\left( \cfrac{r}{s} \right)(x) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ r(6)=3(6)-1\implies r(6)=17~\hfill s(6)=2(6)+1\implies s(6)=13 \\\\[-0.35em] ~\dotfill\\\\ \left( \cfrac{r}{s} \right)(6)\implies \cfrac{r(6)}{s(6)}\implies \cfrac{17}{13}$

7 0

3 years ago

Can get some help plz

Vadim26 [7]

A is the correct answer “V” (16w)(5w)

5 0

3 years ago