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valentinak56 [21]
3 years ago
9

What is the surface area of the triangular prism? A triangular prism. The base is 6 feet by 9 feet. The 2 rectangular sides are

9 feet by 5 feet. The triangular sides have a base of 6 feet and height of 4 feet. _________ square feet
Mathematics
1 answer:
Darina [25.2K]3 years ago
6 0

Answer:

  168 ft²

Step-by-step explanation:

The area of the two triangular bases is ...

  total base area = 2×(triangle area) = 2×(1/2)bh = (6 ft)(4 ft) = 24 ft²

The lateral area of the prism is the product of the triangle perimeter and the height of the prism.

  lateral area = (6 ft + 5 ft + 5 ft)(9 ft) = 144 ft²

The surface area is the sum of the base area and the lateral area:

  surface area = base area + lateral area

  surface area = 24 ft² +144 ft² = 168 ft²

The surface area of the prism is 168 square feet.

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What is the length of a rectangular garden whose perimeter is 32 cm and area 39 cm^2?
kherson [118]

Answer:

Length = 13 cm.

Step-by-step explanation:

If the length is L and the width is W we have the system:

2L + 2W = 32

LW = 39

From the first equation:

L + W = 16 and

L = 16 - W

Substitute this in the second equation:

W(16 - W) = 39

16W - W^2 = 39

W^2 - 16W + 39 = 0

(w - 13)(W - 3) = 0

W = 13, 3

As W < L,  W must be 3  because L = 16-13 = 3 does not make sense.

So Length = 16 - 3 = 13.

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3 years ago
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yawa3891 [41]

Answer:

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Step-by-step explanation:

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1) We have a sequence of 8,13,x

2) To find the number being added every time, we do 13-8. This is 5.

3) This means x would be 13+5

4) 13+5 is 18 so x is 18.

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Use the graph to complete the statement
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Step-by-step explanation:

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fter production, a computer component is given a quality score of A, B, and C. On the average U of the components were given a q
RoseWind [281]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

In the given question some of the data is missing so, its correct question is defined in the attached file please find it.

Let

A is quality score of A

B is quality score of B

C is quality score of C

\to P[A] =0.55\\\\\to P[B] =0.28\\\\\to P[C] =0.17\\

Let F is a value of the content so, the value is:

\to P[\frac{F}{A}] =0.15\\\\\to P[\frac{F}{B}] =0.12\\\\\to P[\frac{F}{C}] =0.14\\

Now, we calculate the tooling value:

\to p[\frac{C}{F}]

using the baues therom:

\to p[\frac{C}{F}] =  \frac{p[C] \times p[\frac{F}{C} ]}{p[A] \times p[\frac{F}{A}] + p[B] \times p[\frac{F}{B}]+p[C] \times p[\frac{F}{C}] }  

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7 0
3 years ago
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