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grandymaker [24]
3 years ago
6

Please help me the picture is above I really need an answer please, I’ll mark as brainliest

Mathematics
1 answer:
sergeinik [125]3 years ago
7 0

Answer:

Option d) The rocket was above 300 feet for approximately 2 seconds.

Explanation:

The rocket was above 300 for both 2 and 4 seconds.

Mark brainliest!!

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Given points A (-3,-4) and B (2,0), points P (-1,-12/5) partitions line AB in the ratio
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\bf \begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;&A&(~ -3 &,& -4~) &#10;&P&(~ -1 &,& -\frac{12}{5}~)&#10;\end{array}&#10;\\\\\\&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;AP=\sqrt{[-1-(-3)]^2+[-\frac{12}{5}-(-4)]^2}&#10;\\\\\\&#10;AP=\sqrt{(-1+3)^2+(-\frac{12}{5}+4)^2}\implies AP=\sqrt{2^2+\left( \frac{8}{5} \right)^2}&#10;\\\\\\&#10;AP=\sqrt{4+\frac{64}{25}}\implies AP=\sqrt{\cfrac{164}{25}}\implies AP=\cfrac{\sqrt{164}}{\sqrt{25}}&#10;\\\\\\&#10;\boxed{AP=\cfrac{2\sqrt{41}}{5}}



\bf \begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;&P&(~ -1 &,& -\frac{12}{5}~) &#10;&B&(~ 2 &,& 0~)&#10;\end{array}&#10;\\\\\\&#10;PB=\sqrt{[2-(-1)]^2+[0-\left(-\frac{12}{5}  \right)]^2}&#10;\\\\\\&#10;PB=\sqrt{(2+1)^2+\left( \frac{12}{5} \right)^2}\implies PB=\sqrt{9+\frac{144}{25}}\implies PB=\sqrt{\cfrac{369}{25}}&#10;\\\\\\&#10;PB=\cfrac{\sqrt{369}}{\sqrt{25}}\implies \boxed{PB=\cfrac{3\sqrt{41}}{5}}

so, let's check the ratio of AP:PB

\bf AP:PB\implies \cfrac{AP}{PB}\implies \cfrac{\frac{2\sqrt{41}}{5}}{\frac{3\sqrt{41}}{5}}\implies \cfrac{2\underline{\sqrt{41}}}{\underline{5}}\cdot \cfrac{\underline{5}}{3\underline{\sqrt{41}}}\implies \cfrac{2}{3}
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3 years ago
Type the missing number that makes these fractions equal: 1 2 = 2
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Answer:

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Step-by-step explanation:

\frac{1}{2}  * 4 = 2

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