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2 years ago

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A square pyramid. The square base has side lengths of 6 inches. The 4 triangular sides have a base of 6 inches and height of 9 i

nches.
Complete the sentences.

The area of the pyramid’s base is in.2.

The pyramid has lateral faces.

The surface area of each lateral face is in.2.

The total surface area is 144 in.2.

1 answer:

Marrrta [24]2 years ago

3 0

Answer:

The area of the pyramid’s base is 36 in².

The pyramid has 4 lateral faces.

The surface area of each lateral face is 27 in².

Step-by-step explanation:

"<u>Lateral</u>" means side, so the lateral faces are <u>triangles</u>.

The <u>base</u> is the bottom, which is a <u>square</u>.

To calculate the <u>area of the base</u>, use the formula for area of a square.

$A_{base} = s^{2}$

$A_{base} = (6 in)^{2}$

$A_{base} = 36 in^{2}$

To calculate the <u>area of a lateral face</u>, find the area of a triangle.

$A_{lateral} = \frac{bh}{2}$

$A_{lateral} = \frac{(6 in)(9 in)}{2}$

$A_{lateral} = \frac{54 in^{2}}{2}$

$A_{lateral} = 27 in^{2}$

In a pyramid, the number of lateral faces is the same as the number of sides in the base. <u>The square base as 4 sides, so there are 4 lateral faces</u>.

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Answer: The surface area of the planter will be 120.5 square feet.

In our shape, we will have 5 sides. The base and the four sides going up from each edge of the base.

The base will be 7 x 5 = 35 square feet.

The front and back side will each be 7 x 0.75 = 5.25 square feet.

The left and right side will each be 5 x 0.75 = 3.75 square feet.

If we add up the 5 faces we get:
35 + 5.25 + 5.25 + 3.75 + 3.75 = 120.5 square feet

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3 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

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