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3 years ago

9

Tyler built a doll house for his sister. The lateral edge of the roof is 13 inches and the dimensions of the base of the house i

s 18 inches x 24 inches. The height of the wall is 10inches. Find the volume of the house.

1 answer:

melamori03 [73]3 years ago

5 0

I think it is 5616 I times 13 18 24

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Solve for x: 3|x-3| 2=14<br> a. No solution<br> b. x=-1,x=8.3<br> c. x=0,x=7<br> d. x=-1,x=7

VMariaS [17]

I am looking on the answers, and there is only one case, when a or b or c or d pass: 3|x-3| + 2 = 14. So I assume, that before two is plus. Then:

3|x-3|+2=14 |minus 2
3|x-3|=12 |divide 3
|x-3|=4

From absolute value definition you've got two ways:

x-3=4 or x-3=-4
x=7 or x=-1

And answer d) passes

3 0

3 years ago

Two number cubes are tossed. What is P(4, 6)?<br><br> O.1/6<br> O.1/12<br> O.1/3<br> O.1/36

Bezzdna [24]

The value of P(4, 6) when the two number cubes are tossed is 1/36

<h3>How to determine the probability?</h3>

On each number cube, we have:

Sample space = {1, 2, 3, 4, 5, 6}

The individual probabilities are then represented as:

P(4) =1/6

P(6) =1/6

The value of P(4, 6) when the two number cubes are tossed is:

P(4, 6) = P(4) * P(6)

This gives

P(4, 6) = 1/6 * 1/6

Evaluate

P(4, 6) = 1/36

Hence, the value of P(4, 6) when the two number cubes are tossed is 1/36

Read more about probability at:

brainly.com/question/24756209

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1 year 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

James invest some money into an account that pays 4% compounded annually if the account is $851.66 after five years how much do

Svetradugi [14.3K]

Answer:

Your moma

Step-by-step explanation:

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3 years ago

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Answer fast please will give brainliest<br> 4 × 10-5 cm=?

Nesterboy [21]

Answer:

35cm

Step-by-step explanation:

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2 years ago

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