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muminat
3 years ago
15

Jennifer wants to wrap a gift box. The gift box has a square base with a side length of 1.2 feet and a height of 1 foot. The cos

t of wrapping paper is $0.50 per square foot.
How much will it cost Jennifer to wrap the gift box?
Mathematics
2 answers:
Karo-lina-s [1.5K]3 years ago
7 0
So the area of the bases top and bottom = 28.8 sq ft. The area of the sides = 4.8 sq ft. add those together it's 33.6 now divide that by 2 and you'll get 16.8$ for Jennifer to wrap the box. 
Klio2033 [76]3 years ago
6 0
The price would be 0.60. It's half of 1.2
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gregori [183]

9514 1404 393

Answer:

  x = 7

Step-by-step explanation:

Using Cramer's rule, the value of x is ...

  x=\dfrac{\left|\begin{array}{cc}2&3\\22&-2\end{array}\right|}{\left|\begin{array}{cc}-1&3\\4&-2\end{array}\right|}

  x = ((2)(-2) -(22)(3))/((-1)(-2) -(4)(3)) = (-4-66)/(2-12) = -70/-10

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2 years ago
How to order -14.4, -14, -14 1/5 from least to greatest
ElenaW [278]
Think of  thermometer the negative degrees are the coldest so are the lowest on the vertical column

first convert -14 1/5 to decimals  =  -14.2

so the order is 

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8 0
2 years ago
A triangular lot has sides of 215m, 185m, and 125m. Find the measures of the angles at its corners
Vlada [557]

Answer:

The measures of the angles at its corners are 59.1\°,35.4\°,85.5\°

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of angle A

Applying the law of cosines

185^{2}= 215^{2}+125^{2}-2(215)(125)cos(A)

2(215)(125)cos(A)= 215^{2}+125^{2}-185^{2}

cos(A)= [215^{2}+125^{2}-185^{2}]/(2(215)(125))cos(A)=0.513953

A=arccos(0.513953)=59.1\°

step 2

Find the measure of angle B

Applying the law of cosines

125^{2}= 215^{2}+185^{2}-2(215)(185)cos(B)

2(215)(185)cos(B)= 215^{2}+185^{2}-125^{2}

cos(B)= [215^{2}+185^{2}-125^{2}]/(2(215)(185))cos(B)=0.81489

B=arccos(0.81489)=35.4\°

step 3

Find the measure of angle C

Applying the law of cosines

215^{2}= 125^{2}+185^{2}-2(125)(185)cos(C)

2(125)(185)cos(C)= 125^{2}+185^{2}-215^{2}

cos(C)= [125^{2}+185^{2}-215^{2}]/(2(125)(185))cos(C)=0.0784

C=arccos(0.0784)=85.5\°

3 0
3 years ago
A study found a relationship between coaching style and points scored in the final two minutes of a basketball game. What is the
harina [27]

Answer:

The final 2 minutes of the basketball game

Step-by-step explanation:

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7 0
2 years ago
Need help finding n forgot how to do this
andreyandreev [35.5K]

Answer:

n  =                                    3\sqrt{2}

There are two best ways to solve this.

using cosine method:

cos(n) =  \frac{adjacent}{hypotenuse}

cos(60) = \frac{adjacent}{6\sqrt{2} }

adjacent,n = cos(60) * 6\sqrt{2}

n  =                                    3\sqrt{2}

using sine method:

sin(n) = \frac{opposite}{hypotenuse}

sin(30)=  \frac{n}{6\sqrt{2} }

n  = sin(30) * 6\sqrt{2}

n = 3\sqrt{2}

There are many ways, not to make it complex, these are the best ways to solve for n. Hope it helps ~

5 0
1 year ago
Read 2 more answers
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