Help ill give brainliest ofc

Mathematics

1 answer:

AlladinOne [14]3 years ago

5 0

Answer:

i think it's 59

Step-by-step explanation:

my bad if u get it wrong or anything

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g100num [7]

I am pretty positive that it would, the circumference of the tube comes out to roughly 6.28 inches, so thee 8 inch wrapping paper should cover it nicely. Now the sheet of paper is only 2 inches longer than the tube, so the paper will only extend one inch above the tubes edge on either end. the tube has a diameter of 2 inches, so the 1 inch extension all around the tube should fold in and match up quite nicely.

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A company sells I shirts for $6 each. They also charge a one time processing fee of $3

Maksim231197 [3]

Answer:

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3 years ago

Three mangos and two apples sell for $1.25. Five mangos and one apple sell for $1.50. What is the cost for each mango

Pepsi [2]

Answer:

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Step-by-step explanation:

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3 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

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3 years ago

Math Question help please, <br> ill give you points ;D thank you

erastovalidia [21]

The answer is A with the slope being 3/2 and the y intercept being -5.

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4 years ago