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3 years ago

A sandbox 12ft. By 14 ft. requires that the sand be spread to a depth of 6 in. How many cubic feet of sand are needed?

Mathematics

2 answers:

Fynjy0 [20]3 years ago

5 0

Answer:

84 ft²

Step-by-step explanation:

Solve for Volume. Volume of a box is:

V = Length (base) x Width (base) x Height (of rectangular prism/square)

Change each measurement to have the same measurement (ft -> in, or vice versa).

Note that 1 ft = 12 in.

6 in = 1/2 ft, because 6/12 = 1/2

Length = 12 ft

Width = 14 ft

Height = 1/2 ft

Solve. Plug in the corresponding number to the corresponding words.

V = 12 x 14 x 1/2

Simplify. Solve.

V = 12 x (14 x 1/2)

V = 12 x (14/2)

V = 12 x 7

V = 84

84 ft² is your answer.

galina1969 [7]3 years ago

3 0

Answer:

1,008

Step-by-step explanation:

You multiply all three numbers to get your answer.

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3 years ago

You place a 16 foot ladder against a building. The angle that the ladder forms with

gulaghasi [49]

Answer:The base of the ladder should be placed so that it is one foot away from the building for every four feet of hight to where the ladder rests against the building. This is known as the 4 to 1 rule.

Step-by-step explanation:

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3 years ago

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

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

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

Then, the minimum point of this function is (3,1) and it corresponds to (A)

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

Then, the minimum point of this function is (0,1) and it corresponds to (C)

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

Then, the minimum point of this function is (2,3) and it corresponds to (B)

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

Then, the minimum point of this function is (2,-2) and it corresponds to (D)

I hope it helps you!

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3 years ago

Fill in the blank. Given O below, you can conclude that OD is congruent to ___________.

Debora [2.8K]

The circle with center O has two chords AC and EF which are of same length 9.07.

OD and OB are the two perpendiculars drawn from the center O to the two chords AC and EF .It represents the distance of the chords from the centre.

The circle theorem states: congruent chords are equidistant from the center.

OD is congruent to OB.

Option A is the right answer.

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3 years ago

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12 out of 15 students play in the school band. In a class of 45 students, how many do NOT play in the band? |

jek_recluse [69]

Answer:36

becuse if you maltiply 15 by 3 it makes 45 and you can maltiply 12 by 3 and you will get your answer

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