3(cx - 7) = 9 the value of x in the equation is __

Cloud [144]

Answer:

Ø

Step-by-step explanation:

Anything that is set to equal x is what is known as an undefined rate of change [slope] because it is a vertical line.

I am joyous to assist you anytime.

8 0

3 years ago

How do you reduce 75/100?

Anestetic [448]

Find the greatest common factor and then divide each number by it.

in this case the gcf for 75 and 100 is 25.

75 divided by 25 = 3
100 divided by 25 = 4

so 75/100 simplified = 3/4

4 0

4 years ago

Read 2 more answers

If the ratio of the measure of the complement of an angle to the measure of its supplement is 1:3, find the measure of the angle

prohojiy [21]

Set up and solve an equation:

3(90-x) = (180-x). Then 270 - 3x = 180 - x, or 270 - 180 = 2x.
90 = 2x, so x = 45 deg.

Is this true? 3(90-45) = (180-45)? If so, x = 45 degrees is correct.

3 0

3 years ago

Fill in the rest of the table

Strike441 [17]

.................................

6 0

3 years ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

3 years ago

3(cx - 7) = 9 the value of x in the equation is ___