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vfiekz [6]
2 years ago
8

A skydiver descended at a constant rate after opening her parachute. Three minutes after she opened her parachute, she had

Mathematics
1 answer:
Reptile [31]2 years ago
3 0

Answer:

1056 feet per a minute

Step-by-step explanation:

math

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The mean of 5 numbers is 12. If four of the numbers are 1, 2, 3, 20, what is the last number?
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Write a system of equations that could be used to solve the situation described below.
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If there are 12 coins under the couch and every coin is either a nickel or a penny and together they are forming 32 cents then the equations which represent the problem will be x+y=12, 0.01x+0.05y=0.32. The correct option is B which is x+y=12, 0.01x+0.05y=0.32.

Given There are 12 coins and total money is 32 cents.

Let the number of pennies be x and the number of  nickels be y.

According to question there are 12 coins so the first equation becomes:

x+y=12.

Then we have been told that together they amount to 32 cents.

We know that 1 penny is 1 cent coin and 1 nickel is 5 cent coin so the equation becomes :

1*x+5*x=32

x+5y=32

converting into dollar

0.01x+0.05y=0.32.

Hence the right equations showing the problem of nickel and pennies are x+y=12, 0.01x+0.05y=0.32.

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HELP FIND PERCENT OF CHANGE AND INCREASE OR DECREASE
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Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
3 years ago
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