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Dmitry [639]
3 years ago
13

One face of a cube-shaped package has an area of 7 5/8 square inches what is the lateral surface area of the package in square i

nches
Mathematics
1 answer:
tamaranim1 [39]3 years ago
6 0

Answer:

232.57 square inches

Step-by-step explanation:

The computation of the  lateral surface area of the package in square inches is shown below;

Let us assume the area of cube is a

Given that

a = 7 \frac{5}{8}\ in^2

As we know that in a cube there is a four faces so the lateral surface area would be

= 4a^2\\\\=4\times 7\frac{5}{8} ^2\\\\= 4\times (\frac{61}{8})^2\\\\= 4\times  \frac{3721}{64} \\\\= 232.57

= 232.57 square inches

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4 Times the sum of a number and 6 is 5 less than 3 times the same number. write an equation and solve for the number
Vladimir [108]

Answer:

-29

Step-by-step explanation:

4× (x+6) = -5. + 3x

4 times the sum of a number is 5 less than 3 times the same number

4×(x+6) = 3x - 5

4x + 24 = 3x - 5

x = -29

3 0
3 years ago
Pls helppp!!! -------
Gelneren [198K]

Answer:

∠ A ≈ 44.42°

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos A = \frac{adjacent}{hypotenuse}  = \frac{AC}{AB} = \frac{5}{7} , then

∠ A = cos^{-1} (\frac{5}{7} ) ≈ 44.42° ( to the nearest hundredth )

5 0
3 years ago
What is the value of -42 +(5-2)(-3)?
SpyIntel [72]

Answer:

=-42+3-3

=-42-3+3

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Step-by-step explanation:

this is the correct answer

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6 0
3 years ago
Solve absolute value equation 2|x-3|-8=2
Semmy [17]

Answer:

x=8, -2

Step-by-step explanation:

2|x-3|-8=2

2|x-3|=10

|x-3|=5                               |x-3|=-5

x-3=5                                x-3=-5

x=8                                      x=-2

7 0
3 years ago
Area triangles? it says show work so pls do
coldgirl [10]

\large{\underline{\underline{\pmb{\sf {\color {blue}{Solution:}}}}}}

Given,

Area = 225 yd²

Base = 30 yd

Height = [To be calculated]

To find:

The height of the given triangle.

We know that, area of a triangle is:

\frac{1}{2}  \times (b \times h) \\  \\  \leadsto \frac{1}{2} \times (30 \times h) \\  \\  \leadsto \frac{1}{2}   \times 30h \\  \\  \leadsto  \frac{1}{ \cancel{2}} \times  \cancel{30}h \\  \\  \leadsto1 \times 15h \\  \\  \leadsto \: h = 15 \: yd

Therefore, the require height is 15 yd.

\green\starProof:

\frac{1}{2}  \times (b \times h) \\  \\  \leadsto \frac{1}{2}  \times (30 \times 15) \\  \\  \leadsto \frac{1}{ \cancel{2}}   \times  \cancel{450} \\  \\  \leadsto1 \times 225  \\  \\  area = 225  {yd}^{2}

\boxed{ \frak \red{brainlysamurai}}

6 0
2 years ago
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