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Andrews [41]
3 years ago
10

Can you please find the surface area!

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
3 0

\text {Area of the triangle } = \dfrac{1}{2} (6) (2) = 6

Area of 2 triangles = 6 x 2 = 12 units²

-

Area of the lateral surface = (3 + 3 + 6) x 7 = 84 units²

-

Total Surface Area = 12 + 84 = 96 units²

-

Answer: 96 units²

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Nataly [62]

Answer:

W is not a subspace.

Step-by-step explanation:

Notice that (-1,2,2) and (-1,5,3) are both elements of W. Because its x coordinate, x = -1, satisfy the condition |x| = 1.

but (-1,2,2) + (-1,5,3) = (-2,7,5) do not belongs to W. Because its first coordinate, x =-2, do not satisfy |x| = 1.  

5 0
3 years ago
Please help will be marked BRAINLIEST!
AysviL [449]

Answer:

shorter base = 6 yd

longer base = 8 yd

area of playground = 42 yd^2

Step-by-step explanation:

The question tells you that the shorter side is equal to the width, which is 6 yds. The bases of the triangles on either side are 1 yd. Since there are two, add 1 twice to 6 (6+1+1). This gives you 8 yd for the longer base. You're right about the area, just use the trapezoid area formula: 1/2h(a+b).

4 0
3 years ago
Last year 858 people attended the Middleton County Fair. This year 1,635 people attended. What was the percent increase from las
Pepsi [2]

Answer:

91%

Step-by-step explanation:

The first step is to calculate the increase

= 1635-858

= 777

Therefore the percent increase can be calculated as follows

= 777/858 × 100

= 0.905× 100

= 90.5 %

= 91%

Hence the percentage increase is 91%

6 0
3 years ago
Which of the following is the radical expression of 4 times d to the three eighths power? (2 points) 4 times the eighth root of
stepan [7]

For this case we have the following expression:

4 ^ {\frac {3} {8}}

By definition of properties of powers and roots we have to:

a ^ {\frac {m} {n}} = \sqrt [n] {a ^ m}

Then, we can rewrite the expression as:

\sqrt [8] {4 ^ 3}

Answer:

4 ^ {\frac {3} {8}} = \sqrt [8] {4 ^ 3}

4 0
3 years ago
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Artist 52 [7]
Choice D
364 centimeters
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3 years ago
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