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MrMuchimi
3 years ago
12

You’re planning to buy a boat , a motor , and some miscellaneous boating equipment. The boat costs $325 , the motor cost $225 ,

and the equipment cost $ 79 if you’ve saved $438 toward these purchases , how much more do you need to save ?
Mathematics
1 answer:
stealth61 [152]3 years ago
7 0
I am not sure what the question is.
You’re planning to buy a boat , a motor , and some miscellaneous boating equipment. The boat costs $325 , the motor cost $225 , and the equipment cost $ 79 if you’ve saved $438 toward these purchases , how much more do you need to save ? 325+225+79=629       629-438= 191
I think its 191
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Suppose the initial height of a Pumpkin is 12 feet and the pumpkin is being launched with a velocity of 61 feet per second. Use
iogann1982 [59]
<h2>Hello!</h2>

The answer is:

The maximum height before landing will be 69.7804 feet.

<h2>Why?</h2>

Since there is no information about the angle of the launch, we can safely assume that it's launched vertically.

So, we can calculate the maximum height of the pumpkin using the following formulas:

y=y_o+v_{o}*t-\frac{1}{2}gt^{2}

v=vo-gt

Where,

y, is the final height

y_o, is the initial height

g, is the acceleration of gravity , and it's equal to:

g=32.2\frac{ft}{s^{2} }

t, is the time.

Now, we are given the following information:

y_{o}=12ft\\\\v=61\frac{ft}{s}

Then, to calculate the maximum height, we must remember that at the maximum height, the speed tends to 0, so, calculating we have:

Time calculation,

We need to use the following equation,

v=vo-gt

So, substituting we have:

v=61\frac{ft}{s}-32.2\frac{ft}{s^{2}}*t\\\\-61\frac{ft}{s}=-32.2\frac{ft}{s^{2}}*t\\\\t=\frac{-61{ft}{s}}{-32.2\frac{ft}{s^{2}}}=1.8944s

We know that it will take 1.8944 seconds to the pumpkin to reach its maximum height.

Maximum height calculation,

Now, calculating the maximum height, we need to use the following equation:

y=y_o+v_{o}*t-\frac{1}{2}gt^{2}

Substituting and calculating, we have:

y=y_o+v_{o}*t-\frac{1}{2}gt^{2}

y=12ft+61\frac{ft}{s}*1.8944s-\frac{1}{2}32.2\frac{ft}{s^{2}}*(1.8944s)^{2}

y=12ft+61\frac{ft}{s}*1.8944s-\frac{1}{2}32.2\frac{ft}{s^{2}}*(1.8944s)^{2}\\\\y_{max}=12ft+115.5584ft-16.1\frac{ft}{s^{2}}*(3.5887s^{2})\\\\y_{max}=127.5584ft-57.7780ft=69.7804ft

Hence, we have that the maximum height before the landing will be 69.7804 feet.

Have a nice day!

3 0
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RoseWind [281]

Answer: B

Step-by-step explanation:

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tankabanditka [31]

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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