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2 years ago

Solve for x: −3x + 3 < 6 x > −1 x < −1 x < −3 x > −3

Mathematics

2 answers:

Misha Larkins [42]2 years ago

7 0

Answer:

x > -1

Step-by-step explanation:

Ad libitum [116K]2 years ago

7 0

$\bf{-3x+3 < 6}$

Subtract 3 from both sides.

$\bf{-3x < 6-3 }$

Subtract 3 from 6 to get 3.

$\bf{-3x < 3}$

Divide both sides by -3. Since −3 is < 0, the inequality direction is changed.

$\bf{x > \dfrac{3}{-3} }$

Divide 3 by -3 to get -1.

$\bf{x > -1 \ \ \to \ \ \ Answer }$

Therefore, the correct option is the first.

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3 years ago

*30 POINTS* In a group of 53 children, 22 had a dog, 18 had a cat, and 6 had both a dog and a cat. How many children had neither

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Answer:

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3 years ago

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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

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STatiana [176]

Answer:

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I hope you find this helpful.

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3 years ago

Read 2 more answers

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Katen [24]

Answer:

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3 years ago