All categories

3 years ago

What is the volume of the pyramid ? 8ft 7ft 7ft

Mathematics

2 answers:

Nuetrik [128]3 years ago

7 0

Answer:

V= 392 ft^3

Step-by-step explanation:

We know that the volume V of a pyramid is one-third the area of the base B times the height h.

If h= 8ft and areabase = 7ft x 7ft = 49 ft^2.

Then, the Volume V is:

V = areabase x height = 49 ft^2 x 8ft = 392 ft^3

So the solution is option D.

amid [387]3 years ago

7 0

Answer:

Volume of pyramid is:

$130\dfrac{2}{3}\ ft^3$

Step-by-step explanation:

We know that the volume V of a pyramid is one-third the area of the base times the height h.

First we find the area of base let it be A.

A=7×7 ft²

A=49 ft²

Volume of pyramid=

$\dfrac{1}{3}\times A\times h$

Volume=

$\dfrac{1}{3}\times 49\times 8$

$= \dfrac{392}{3}\\ \\=130\dfrac{2}{3}\ ft^3$

Hence, Volume of pyramid is:

$130\dfrac{2}{3}\ ft^3$

You might be interested in

What is the equation in vertex form of a parabola with vertex (-2, -2) that opens downward and is stretched by a factor of 5?

Galina-37 [17]

Answer:

D). -5(x + 2)^2 - 2.

Step-by-step explanation:

Parabola which Opens down and vertex (-2, -2) is :

f(x) = a(x + 2)^2 - 2 where a is negative

Stretch by factor 5 gives:

-5(x + 2)^2 - 2.

5 0

2 years ago

Wht are the next three terms in the sequence -3 6 15 24

kogti [31]

6--3=9
15-6=9,
so
24+9=33,
33+9=42,
42+9=51
33,42,51

4 0

3 years ago

Read 2 more answers

There were 16 bald eagles in an area some years later there were 28 bald eagle‘s in the area what is the percent game

ch4aika [34]

........................

3 0

2 years ago

An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline

vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have a seat, we have

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 = 5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

$z = \frac{x-np}{\sqrt{np(1-p)} } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} }$ = - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = $\sqrt{npq}$ = 2.233

P (x≤100) = P $(z$ = P(z<0.34) = 0.63307.

6 0

3 years ago

Type the correct answer in each box. Use numerals instead of words.

lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17 eq(1)

y = -1/2x eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a = 5/4, b =0 and c = -17

$x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}$

Finding value of y:

y = -1/2x

$y=-1/2(\frac{\pm2\sqrt{85}}{5})$

$y=\frac{\pm\sqrt{85}}{5}$

System A has 4 real solutions.

System B

y = x^2 -7x + 10 eq(1)

y = -6x + 5 eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-1 and c =5

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}$

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9 eq(1)

8x - y = -17 eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17

8(-2) - y = -17

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

4 0

3 years ago

Read 2 more answers