Please help me with this i don’t understand

Mathematics

1 answer:

nevsk [136]2 years ago

4 0

Answer:

20.12 ft

Step-by-step explanation:

if you take 6^2 and subtract that from 21^2 then you will get your answer after finding the square root of the answer that the calculator gives you.

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Is there a relationship between the distance and the sum? Is there a relationship between the distance and the difference

liraira [26]

The distance is always the opposite of the difference. The distance is exactly the difference. The distance is the absolute value of the difference. The distance is not related to difference.

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3 years ago

Can someone help me?! i’m stuck between two.

QveST [7]

6 feet.
9.6/8= 1.2, so 1.2*5=6.

6 0

2 years ago

what decimal point is between 1.5 and 1.7 and i know its 1.6 but on my homework it wont say 1.6 it says 1.625 , 1.25 , 1.75 , 1.

Eddi Din [679]

It doesn't matter what's between it as long as it is within that range, if that makes any sense.

1.25 isn't.
1.75 isn't.
1.83 isn't.

1.625 would be your answer. :)

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2 years ago

A man Buy television set at a price exclusive of sales tax for $2124 if sales tax of 12% is changed how much money does the man

Fed [463]

Answer:

2378.88

Step-by-step explanation:

12/100 x 2124

=254.88

2124+254.88=answer

6 0

2 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago