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2 years ago

An rectangular water tank has a base area of 56 sq. ft. and a height of 4 ft. Find its volume in cubic feet.

Mathematics

1 answer:

ira [324]2 years ago

7 0

Answer: 224 cubic feet

Step-by-step explanation: If the base of the tank is 56 sq feet in area then the dimensions could be, (2,28) (4,14) (7,8) for example. However, it doesn't really matter. It's only important that the information we know is that it is area and there are two dimensions. Since volume is determined by base x width x height, we do 56 x 4. Or using one of the factored pairs that works, for example, 8 x 7 x 4 = 224.

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What is the solution for −40 = 10c? A. c = 1 B. C = -1 C. c = -4 D. c = 4

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Answer:

C. c = -4

Step-by-step explanation:

you must get c by itself. you have to divide both sides by 10. this gets rid of the 10c and keeps both sides equal

-40/10 = c

c = -4

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3 years ago

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2 years ago

Read 2 more answers

The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?

Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$A(-6, 2)) = (x_1, y_1)$

$B(2, 6) = (x_2, y_2)$

$AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}$

$AB = \sqrt{(8)^2 + (4)^2}$

$AB = \sqrt{64 + 16} = \sqrt{80}$

$AB = 8.9$ (nearest tenth)

Distance between B(2, 6) and C(7, 1):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(2, 6) = (x_1, y_1)$

$C(7, 1) = (x_2, y_2)$

$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

$BC = \sqrt{(5)^2 + (-5)^2}$

$BC = \sqrt{25 + 25} = \sqrt{50}$

$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

$CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}$

$CD = \sqrt{(-4)^2 + (-6)^2}$

$CD = \sqrt{16 + 36} = \sqrt{52}$

$CD = 7.2$ (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

$DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

$DA = \sqrt{(-9)^2 + (7)^2}$

$DA = \sqrt{81 + 49} = \sqrt{130}$

$DA = 11.4$ (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0

3 years ago

In how many ways can 6 competitors finish a race assuming there are no ties?In how many ways can we sit n people on r < n cha

VLD [36.1K]

Answer: a) 720, b) $^nC_r=\dfrac{n!}{(n-r)!r!}$ c) 336, d) 30240 e) 120.

Step-by-step explanation:

Since we have given that

a) Number of competitors = 6

We need to find the number of ways that 6 competitors can finish a race and there are no ties.

So, Number of ways would be

$6!=720$

b) number of people = n

Number of chairs = r (r<n)

so, the number of ways we can sit n people on r<n chairs is given by

$^nC_r=\dfrac{n!}{(n-r)!r!}$

c) If there are three awards :

Gold, silver, bronze.

We need to find the number of ways that these medals can be awarded.

Number of ways is given by

$^8P_3\\\\=336$

d) Number of 5-digit zip codes are possible is given by

$^{10}P_5=30240$

e) Number of poeple = 10

Number of people required in committee = 3

Number of different 3 person committee that are selected from 10 people is given by

$^{10}C_3=120$

Hence, a) 720, b) $^nC_r=\dfrac{n!}{(n-r)!r!}$ c) 336, d) 30240 e) 120.

6 0

3 years ago