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Kisachek [45]
3 years ago
15

A parachutist descends 52 feet in 4 seconds. Express the rate of the parachutists change in height as a unit rate.

Mathematics
1 answer:
rosijanka [135]3 years ago
5 0

Answer:

the answer is 13 per one sec

Step-by-step explanation:

4 times 13 is 52 so every one sec its 13

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24 or 144.... figure it out bud

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Length x width x height=

6 0
3 years ago
You purchase a new motor scooter for $3,350. You put a 10% down payment and $108.08 per month on a 36 month purchase plan. Use t
nevsk [136]
Use the formula of the present value of annuity ordinary through GoogleWhat you have here is a loan payment of $108.08 with a present value of $3015 (the $3350 minus the 10% down payment) and a future value of zero with monthly compounding over 36 months
I got
R=0.173906
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7 0
3 years ago
A 4m ladder rest against a vertical wall with its foot 2m from the wall.How far up the wall does the ladder reach?
adell [148]

A) How high up the wall does it reach?

Use the Pythagorean theorem

Height^2 + Base^2 = Hyp^2

H^2+ 2^2 = 4^2         Subtract the 2^2 from both sides

H^2 = 4^2 -2^2          Multiply the square roots of both the numbers

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Hope this helps

4 0
3 years ago
The value of the definite integral / 212x – sin(x) dx / 122x-sin(x) is
blondinia [14]

Hi there!

\boxed{= 70 + cos(12) - cos(2) \approx 71.26}

\int\limits^{12}_{2} {x-sin(x)} \, dx

We can evaluate using the power rule and trig rules:

\int x^n = \frac{x^{n+1}}{n+1}

\int x = \frac{1}{2}x^{2}

\int -sin(x) = cos(x)

Therefore:

\int\limits^{12}_{2} {x-sin(x)} \, dx = [\frac{1}{2}x^{2}+cos(x)]_{2}^{12}

Evaluate:

(\frac{1}{2}(12^{2})+cos(12)) - (\frac{1}{2}(2^2)+cos(2))\\= (72 + cos(12)) - (2 + cos(2))\\\\= 70 + cos(12) - cos(2) \approx 71.26

3 0
3 years ago
How would you classify the number 121 ?<br>​
Maslowich

Answer:

\boxed{121\:  is \:  a \: perfect \: square}

Step-by-step explanation:

the \:  \sqrt[3]{121}  \: is \: not \: a \: perfect \: cube \\ but \\ the \:  \sqrt{121}  = 11 \to \: is \: a \: perfect \: square \\

5 0
3 years ago
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