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Setler79 [48]
3 years ago
15

Titus wants to fence a square area of his yard for his dogs if you wants the area of the square to be a 121 square feet what doe

s the perimeter of the square need to be
Mathematics
2 answers:
Dmitrij [34]3 years ago
5 0

A square of area 121 ft² will have a side dimension of √121 ft = 11 ft. The perimeter is the length of 4 sides, so is 44 ft.

Ne4ueva [31]3 years ago
3 0

Answer: 44 feet.

Step-by-step explanation:

Let s be the side of square area .

Given : Titus wants to fence a square area of his yard for his dogs if you wants the area of the square to be a 121 square feet.

The area of square is given by :-

A=s^2

i.e. s^2=121

Taking square root on both sides, we get

s=11

The perimeter of square is given by :-

P=4s

i.e. P=4(11)=11\text{ feet}

Hence, the perimeter of the square need to be 44 feet.

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Answer:

30

Step-by-step explanation:

Add all of them

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3 years ago
Enter the terms and the coefficients of the expression. −80 + 5p − 8z The terms are . The coefficients are .
9966 [12]

Answer:

Step-by-step explanations:

Hi love ,

So the terms is :

<u>-80 , 5p , -8z </u>

and the coefficients is :

<u>5 , -8</u>

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8 0
3 years ago
What is the reciprocal of 2 and 3/8
Leviafan [203]
To find the reciprocal of a number or fraction, just flip the numerator and denominator. Whole numbers are always over 1. 

 2 = \frac{2}{1}

The reciprocal of \frac{2}{1} =  \frac{1}{2} or .5 (one-half)

The reciprocal of \frac{3}{8} =  \frac{8}{3}
8 0
3 years ago
Read 2 more answers
Let f be the function defined as follows:1. If a = 2 and b = 3, is f continuous at x = 1? Justify your answer.2. Find a relation
Arada [10]

Answer:

1. not continuous, as the function definitions deliver different function values at x=1 when approaching this x from the left and from the right side.

2.

2 = a + b

3.

0 = 2a + b

4.

a = -2

b = 4

Step-by-step explanation:

the function is continuous at a specific point or value of x, if the f(x) = y functional value is the same coming from the left and the right side at that point.

1. that means that for x=1

3 - x = ax² + bx

so,

3 - 1 = a×1² + b×1 = a + b

2 = a + b

we have to use a=2 and b=3

2 = 2 + 3 = 5

2 is not equal 5, so the assumed equality is false, so the function is not continuous there.

2. point 1 gave us already the working relationship between a and b.

2 = a + b

only if that is true, is the function continuous at x=1.

3. now for x=2

5x - 10 = ax² + bx

5×2 - 10 = a×2² + b×2 = 4a + 2b

10 - 10 = 4a + 2b

0 = 4a + 2b

0 = 2a + b

4. to find a and b to be continuous at both locations x=1 and x=2 both expressions in a and b must apply.

so, they establish a system of 2 equations with 2 variables.

2 = a + b

0 = 2a + b

a = 2 - b

0 = 2×(2-b) + b = 4 - 2b + b = 4 - b

b = 4

therefore

a = 2 - 4 = -2

5. I cannot draw a graph here.

just use now the function

3 - x, x < 1

‐2x² +4x, 1 <= x < 2

5x - 10, x >= 2

8 0
3 years ago
State the order and type of each transformation of the graph of the function
Zina [86]

Answer:

steps below (C)

Step-by-step explanation:

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3 years ago
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