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liubo4ka [24]
3 years ago
13

What is the volume, in cubic units, of a pyramid of height 8 and a square base of length 15?

Mathematics
2 answers:
AURORKA [14]3 years ago
7 0

Answer:

900 cuboc unit

Step-by-step explanation:

Volume of pryamid= 1/3× area× height

= 1÷3×15^15×8

= 900cubic unit

Paraphin [41]3 years ago
6 0

For this case we have by definition, that the area of a square base pyramid is given by:

V = \frac {L ^ 2 * h} {3}

Where:

L: It's the side of the square base

h: It's the height of the pyramid

We have thatL = 15 \ units, then L ^ 2 = 15 ^ 2 = 225 \ units ^ 2

Substituting:

V = \frac {225 * 8} {3}\\V = 600 \ units ^ 3

Finally, the volume of the pyramid is 600 \ units ^ 3

Answer:

600 \ units ^ 3

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A rectangle has a perimeter of 204 feet. It's length is six feet longer than twice it's width. If L stands for the length of the
Ratling [72]

Answer:

L = 70

W = 32

Step-by-step explanation:

P = 2L + 2W

P 204 feet

W = x

L  = 2x + 6

P = 2L + 2W

P = 2*(2x + 6) + 2*x

P = 4x + 12 + 2x

P = 6x + 12

But we know that P = 204  So 6x + 12 = 204

6x + 12 = 204                    Subtract 12 from both sides

6x + 12 - 12 = 204 - 12       Collect like terms on the right

6x = 192                             Divide by 6

x = 192/6

x = 32

Answer

W = 32

L = 2*W + 6

L = 2*32 + 6

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L = 70


7 0
3 years ago
The electric-vehicle manufacturing company Tesla estimates that a driver who commutes miles per day in a Model S will require a
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Answer:

Following are the solution to the given points:

Step-by-step explanation:

Please find the complete question in the attached file.

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 In point a:

so, pdf of x =

\to f(x)= \frac{1}{120-90}  = \frac{1}{30} \\\\\to 90 < x

In point b:

To find

\to P(x

In point c:

\to P(x

In point d:

\to P(95< x< 110)

= \int^{110}_{95} \frac{1}{30} \ dx\\\\= \frac{110-95}{30} \\\\= \frac{15}{30} \\\\= \frac{1}{2}

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What is 46.2 x 10 negative 2=
Nutka1998 [239]

The Answer is -924.

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3 years ago
A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f
mestny [16]

Answer:

The value is W= 2640 \  ft \cdot lb

Step-by-step explanation:

From the question we are told that

The weight of the bucket is F =  5 lb

The depth of the well is x_1 =  60 \ ft

The weight of the water is W_w  =  42 lb

The rate at which the bucket with water is pulled is v  = 1.5 \  ft/s

The rate of the leak is r = 0.15 lb/s

Generally the workdone is mathematically represented as

W =  \int\limits^{x_1}_{x_o} {G(x)} \, dx]

Here G(x) is a function defining the weight of the system (water and bucket ) and it is mathematically represented as

G(x) =  F  +  (W_w- Ix)

Here I is the rate of water loss in lb/ft mathematically represented as

I  = \frac{r}{v}

=> I  = \frac{0.15 }{1.5 }

=> I  = 0.1

So

G(x) =  5  +  (42- 0.1x)

=> G(x) =  47- 0.1x)

So

W =  \int\limits^{60}_{0} {47- 0.1x} \, dx]

=> W =  [47x - \frac{0.1x^2}{2} ]|\left 60} \atop {0}} \right.

=> W= [47(60) - 0.05(60)^2]

=> W= 2640 \  ft \cdot lb

7 0
3 years ago
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