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3 years ago

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Carlie is painting a trunk that measures 40 inches wide, 30 inches tall, and 60 inches long. Which is the correct unit of measur

ement to use to figure out the amount of paint Carlie needs to cover the complete trunk? A. in. B. in3 C. in2

1 answer:

HACTEHA [7]3 years ago

6 0

Answer:

in2 because it is surface area

Step-by-step explanation:

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Willy says the slope of the graph is -1/3.what error did she make when finding the slope

Nadusha1986 [10]

Answer:

you did not add a picture or any other over fractions.

Step-by-step explanation:

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3 years ago

a bowl contains 4 peaches and 4 apricots. Maine randomly selects one, puts it back, and randomly selects another. what is the pr

almond37 [142]

1/4 or 25% because...

The probability that Maine picks an apricot is 4/8

The probability that Maine picks an apricot after putting the first apricot back is still 4/8

4/8 x 4/8 = 16/64 = 1/4

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2 years ago

2. Multiply.<br><br> (3x + 2)(5x^2 - 2x + 6)

alukav5142 [94]

15x^3+4x^2+14x+12
Multiply 3x by every value in the second bracket. Then multiply 2 by every value in the second bracket. Add like terms. (Only answers with the same square are like terms.)

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2 years ago

Plz help me well mark brainliest If Correct

Masja [62]

You would have to begin at (2,4) to reach (5,6)

6 0

3 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

3 years ago