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3 years ago

Calculate the volume of the figure.

Mathematics

2 answers:

Margarita [4]3 years ago

8 0

Answer:

Step-by-step explanation:

Formula

Volume = L * w * h

Givens

L = 7 feet

w = 4 feet

h = 5 feet

Solution

V = L * w * h Substitute values into the formula

V = 7 * 4 * 5

V = 140 cubic feet.

Answer

140 ft^3

Effectus [21]3 years ago

6 0

Answer:

D) 140 cubic feet

Step-by-step explanation:

The given information is,

→ Length (l) = 7 ft

→ Breadth (b) = 4 ft

→ Height (h) = 5 ft

Formula we use,

→ Volume of cuboid = l × b × h

Let's solve the problem,

→ l × b × h

→ 7 × 4 × 5

→ [ 140 ft³ ]

Hence, the volume is 140 ft³.

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ratelena [41]

Answer:

The answer to your question is: is = 14 m/s

Step-by-step explanation:

Data

acceleration = 2 m/s² = a

final speed = 20 m/s = sf

time = 3 seconds = t

initial speed = ? = is

Formula

sf = is + at

is = sf - at

Substitution

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3 years ago

This is a tough one :/<br> If f(x) = -x + 7 and g(x) = radical of x– 3,<br> what is (f º g)(4)

Jlenok [28]

Answer:

Step-by-step explanation:

To solve, first plug in 4 as your x value for your g(x) equation.

$g(4)=\sqrt{4-3} \\g(4)=\sqrt{1} \\g(4)=1$

Next, plug in the value of g(4) into your f(x) equation for x.

$f(1)=-1+7\\f(1)=6$

6 0

3 years ago

For the function y=-1+6cos(2pi/7(x-5)) what is the minimum value?

s344n2d4d5 [400]

Answer:

The minimum value of y is -7

Step-by-step explanation:

Given: $y=-1+6\cos(\dfrac{2\pi}{7}(x-5))$

We are given a cosine function.

We need to find minimum value of function.

$y=-1+6\cos(\dfrac{2\pi}{7}(x-5))$

Here y is depends on cosine function.

y is minimum when cosine is minimum because y is directly proportional to cosine function.

Minimum value of cos x is -1

Therefore, For minium value of y

$\cos(\dfrac{2\pi}{7}(x-5))=-1$

Minimum value of y

$y=-1+6(-1)$

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Hence, The minimum value of y is -7

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4 years ago

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ICE Princess25 [194]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

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Vlada [557]

Answer:

D, F.

Step-by-step explanation:

D and F contain x^2.

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