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3 years ago

I NEED HELP ON THIS ONE TOO SAME THING ILL GIVE BRAINLIEST

Mathematics

1 answer:

Readme [11.4K]3 years ago

3 0

Answer:

449 sq ft

Step-by-step explanation:

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Solve for the variable x.

n200080 [17]

Answer:

x = 84/5 = 16.8

Step-by-step explanation:

The angle bisector theorem said that

21/15 = x/12

==> 7/5 = x/12

==> x = 84/5

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3 years ago

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igomit [66]

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4 years ago

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What is f(x)= (x-3)(x+4) in standard form?

ANEK [815]

Answer:

The answer for this problem is x^2+x-12

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3 years ago

PLEASE HELP ME WITH THIS 17 points!! PLEASEEEE IM BEGGING YOU!!!

slega [8]

Answer:

x = 2

Step-by-step explanation:

Step 1: Convert to math

5x + 1 = 3x + 5

Step 2: Solve for <em>x</em>

Subtract 4x on both sides: 2x + 1 = 5
Subtract 1 on both sides: 2x = 4
Divide both sides by 2: x = 2

Step 3: Check

<em>Plug in x to verify it's a solution.</em>

5(2) + 1 = 3(2) + 5

10 + 1 = 6 + 5

11 = 11

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3 years ago

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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

4 years ago