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SVEN [57.7K]
2 years ago
12

Find the surface area of a square pyramid with a base length of 24 cm and a height of 16 cm. Explain how you solved the problem.

Mathematics
1 answer:
icang [17]2 years ago
6 0

Answer:

the answer b

Step-by-step explanation:

[surface area]=[area of 4 triangles sides]+[area of square base]

[area of square base]=24*24--------> 576 cm²

[area of one triangles side]

l=slant height

l²=16²+12²-------> l²=400-----------> l=20 cm

[area of one triangles side]=24*20/2--------> 240 cm²

[area of 4 triangles sides]=4*240----------> 960 cm²

[surface area]=[960]+[576]--------> 1536 cm²

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ruslelena [56]

Step-by-step explanation:

27 \times  \frac{1}{ {3}^{3} }

= 27 \times  \frac{1}{27}

= 1

7 0
2 years ago
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Find the distance of a line given the following points. R(3, 1), W(-1, -2)
olga_2 [115]

Answer:

5 units

Step-by-step explanation:

±√(- 2 - 1)^2 + (- 1 - 3)^2 = ±5 (rej - 5 since distance > 0)

Apply distance formula

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7 0
2 years ago
the following image is a trapazoid.if XW is congruent YZ and measaure of m = 72 degrees. calculate the measure of X
AnnyKZ [126]

Answer:

opposite angle of trapezoid is supplementary

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is your answer

6 0
3 years ago
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Is my answer right? the other options are <br><br> A.0.33<br> B.0.42<br> C.0.58<br> D.0.67
Fittoniya [83]
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4 0
3 years ago
Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = x − ln 8x, [1/2, 2]
insens350 [35]
The given function is 
f(x) =  x - ln(8x), on the interval [1/2, 2].

The derivative of f is
f'(x) = 1 - 1/x
The second derivative is 
f''(x) = 1/x²

A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0  => 1/x = 1  => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079

The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2)  = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.

Answer: 
Minimum value  = 1 - ln(8)
Maximum value = 2 - ln(16)


4 0
3 years ago
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