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natulia [17]
3 years ago
15

What number does x stand for in this equation?

Mathematics
2 answers:
seropon [69]3 years ago
8 0

Answer:

c. -3

Step-by-step explanation:

- 7x + 6= 27 [subtract 6 from both sides]

-7x = 21 [divide by -7 to isolate x]

x = -3

anyanavicka [17]3 years ago
6 0

Answer:

The answer is C

Pay attention, here's why.

Step-by-step explanation:

Let's solve your equation.

−7x+6=27

Step 1: Subtract 6 from both sides.

−7x+6−6=27−6

−7x=21

Step 2: Divide both sides by -7.

−7x /−7 = 21/ −7

x=-3

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Check the picture below.

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\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:

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