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baherus [9]
2 years ago
9

HELP!!!

Mathematics
1 answer:
Naddika [18.5K]2 years ago
5 0

Check the picture below, so it reaches the maximum height at the vertex, let's check where that is

h(t)=64t-16t^2+0 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 64}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (64)^2}{4(-16)}\right)\implies \stackrel{maximum~height}{(2~~,~~\stackrel{\downarrow }{64})}

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Marysya12 [62]

These are 3 questions and 3 answers.

Question 1.

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Explanation:

1) Equation A

\sqrt{x^2+3x-6}=\sqrt{x+2}\\ \\  x^2+3x-6=x+2\\ \\  x^2+2x-8=0\\ \\  (x+4)(x-2)=0\\ \\  x=-4 ; x =2

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2) Equation B.

Since it is a cube root, it is defined for any (negative, zero, and positive) values of x.

Question 2.

Answer: cubing both sides once.

Explanation:

\sqrt[3]{5x-2} -\sqrt[3]{4x} =0\\ \\  \sqrt[3]{5x-2} =\sqrt[3]{4x}\\  \\  (\sqrt[3]{5x-2} )^3=(\sqrt[3]{4x})^3\\  \\  5x-2=4x\\ \\  5x-4x=2\\ \\  x=2

3) Question 3.

Answer: c - 2 = 25 + c + 10√c

Explanation:

\sqrt{c-2} -\sqrt{c} =5\\ \\ \\  \sqrt{c-2} =\sqrt{c} +5\\ \\  (\sqrt{c-2})^{2}  =(\sqrt{c}+5)^2\\  \\  c-2=c+10\sqrt{c} +25\\ \\  c-2=25+c+10\sqrt{c}

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Answer:

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