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aleksklad [387]
2 years ago
10

LeAnn is purchasing.giftwrap for a box that measures 10 inches long, 6 inches wide, and 6 inches tall. Calculate the total area

of the box that will be covered in.giftwrap. Show all of your work for full credit.
Mathematics
2 answers:
tigry1 [53]2 years ago
6 0

Answer:

the answers is 312 in²

Step-by-step explanation:

hope this help

choli [55]2 years ago
3 0

Answer:

\boxed{312~in^2}

Step-by-step explanation:

(P) Long = 10 in

(L) Wide = 6 in

(T) Tall = 6 in

.

So, Total area is

= 2(P\times L + P \times T + L\times T)

= 2(10\times 6 + 10 \times 6 + 6\times 6)

= 2(60 + 60 + 36)

= 2(156)

= 312~in^2

.

Happy to help :)

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ANSWER:
5a+2b

add the like terms together

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If 4y + 4x = 12, what is y in terms of x?
Hoochie [10]

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Y=-x+3

Step-by-step explanation:

First move the the x term to one side. Isolate the y term. Then because the coefficient is 4 in the y term divide the whole equation by 4. Then you get your answer.

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The length of a rectangular piece of steel in a bridge is 4 meters less than triple the width. The perimeter is 64 meters. Find
MaRussiya [10]
Consider the width as x.
Width = x
Length = 3x-4
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6 0
3 years ago
9. The volume of a triangular prism is 30 cm. Each triangular face has an area of 4 cm. How long is the prism?
stealth61 [152]

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12

Step-by-step explanation:

3 0
3 years ago
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A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by y =
GaryK [48]

Answer:

The distance travelled by the kite is 122.8 ft ( approx )

Step-by-step explanation:

Here, the given function,

y=150-\frac{1}{40}(x-50)^2

Differentiating with respect to x,

y'=-\frac{1}{20}(x-50)

∵ arc length of a curve is,

L=\int_{a}^{b} \sqrt{1+y'^2}dx

Where, y shows the height of the curve for a ≤ x ≤ b,

Thus, the arc length of the given curve is,

L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx

Put -\frac{1}{20}(x-50)=tan\theta

\implies -dx=-20 sec^2\theta d\theta

\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta

=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta

=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta

By integration by parts,

=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}

If x = 80, tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}

sec \theta = \frac{\sqrt{13}}{2}

\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}

sec \theta = \frac{\sqrt{29}}{2}

Thus, the length of the curve is,

=-10(\frac{\sqrt{13}}{2}(-\frac{3}{2}) +ln|\frac{13}{2}-\frac{3}{2}|) + 10(\frac{5\sqrt{29}}{4} + ln |\frac{29}{2} + \frac{5}{2} |)

\approx 122.8\text{ feet}

8 0
3 years ago
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