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guajiro [1.7K]
3 years ago
10

Which is the area of a rectangle with a length of 21 feet and a width of 15 feet

Mathematics
2 answers:
muminat3 years ago
6 0

Answer:

315 ft^2

Step-by-step explanation:

SpyIntel [72]3 years ago
5 0

Answer: 315\ ft^2

Step-by-step explanation:

In order to calculate the area of a rectangle, you can use the following formula:

A=lw

Where "l" is the lenght of the rectangle and "w" is the width.

In this case you know that this rectangle has a length of 21 feet and a width of 15 feet. Then:

l=21\ ft\\w=15\ ft

Therefore, you can substitute these values into the formula, getting that the area of this rectangle is:

A=(21\ ft)(15\ ft)\\\\A=315\ ft^2

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6/25

Step-by-step explanation:

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F(x)=4(x+2)(x+2)-6 in standard form
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Answer:

4x^2 + 4x -2

Step-by-step explanation:

Use FOIL to solve (x+2) (x+2)

4(x^2 + 4x +4) - 6

4x^2 + 4x -2

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How do i graph 3/2 on the number line?<br>​
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by drawing a number line then simpifly 3/2 than mark it down

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3 years ago
Required information Skip to question NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to re
pogonyaev

The result for the given 14 length bit string is-

(a) The bit strings of length 14 which have exactly three 0s is 2184.

(b) The bit strings of length 14 which have same number of 0s as 1s is 17297280.

(c) The bit strings of length 14 which have at least three 1s is 4472832.

<h3>What is a bit strings?</h3>

A bit-string would be a binary digit sequence (bits). The size of the value is the amount of bits contained in the sequence.

A null string is a bit-string with no length.

The concept used here is permutation;

ⁿPₓ = n!/(n-x)!

Where, n is the total samples.

x is the selected samples.

(a) Because there are exactly three 0s, its other digits have always been one, hence the total of permutations is equal to;

n = 14 and x = 3 digits.

¹⁴P₃ = 14!/(14-3)!

¹⁴P₃ = 2184.

Thus, the bit strings of length 14 which have exactly three 0s is 2184.

(b) Because it is a bit string with 14 digits and it can only have digits 0 or 1, we must choose 7 0s and the remaining 7 1s, hence the number possible permutations is equal to;

n = 14 and x = 7 digits.

¹⁴P₇ = 14!/(14-7)!

¹⁴P₇ = 17297280.

Thus, the bit strings of length 14 which have same number of 0s as 1s is 17297280.

(c) The answer is identical to problem a, but the rest of a digits could be either 0 or 1, hence it must be doubled by 2¹¹ because there are 11 digits, each of which can be one of two options;

= 2184×2¹¹

= 4472832

Thus, he bit strings of length 14 which have at least three 1s is 4472832.

To know more about permutation, here

brainly.com/question/12468032

#SPJ4

The correct question is-

How many bit strings of length 14 have:

(a) Exactly three 0s?

(b) The same number of 0s as 1s?

(d) At least three 1s?

3 0
1 year ago
How do you solve 0.8=6-2.5x
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X=2.08

To solve you need to first get x by itself. To do that you want to subtract 6 from both sides (get 6 away from x and put on other side) which will leave you with -5.2=-2.5x next isolate x by dividing -2.5 on both sides to then get 2.08=x
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