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N76 [4]
3 years ago
14

A pentagon has 3 congruent sides and 2 other congruent sides. The perimeter of the pentagon is 36 centimeters. The three long co

ngruent sides are 2 centimeters longer than the two shorter congruent sides. Let x = length of a short side Let y = length of a long side The system of equations can be used to represent the situation. y = x + 2 2x + 3y = 36 What is the length of one of the shorter congruent sides?
A. 2 centimeters
B. 6 centimeters
C. 8 centimeters
D. 17 centimeters
Mathematics
2 answers:
Charra [1.4K]3 years ago
6 0

Answer:

B 6 centimeters

Step-by-step explanation:

y = x + 2

2x + 3y = 36

substitute y = x+2  into the second equation

2x + 3 (x+2) = 36

distribute

2x + 3x+ 3*2 = 36

combine like terms

5x+6 = 36

subtract 6 from each side

5x +6-6 = 36 -6

5x = 30

divide by 5

5x/5 = 30/5

x = 6

the short side is x =6

the longer side is

y= x+2

y = 6+2

y =8

12345 [234]3 years ago
3 0

Answer:

b

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

1/6 / 4

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2 years ago
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Suppose a rectangular pasture is to be constructed using 1 2 linear mile of fencing. The pasture will have one divider parallel
timama [110]

Answer:

\displaystyle A=\frac{1}{192}

Step-by-step explanation:

<u>Maximization With Derivatives</u>

Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.

The amount of fence needed to enclose the external and the internal divisions is

P=4x+3y

We know the total fencing is 1/2 miles long, thus

\displaystyle 4x+3y=\frac{1}{2}

Solving for x

\displaystyle x=\frac{\frac{1}{2}-3y}{4}

The total area of the pasture is

A=x.y

Substituting x

\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y

\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}

Differentiating with respect to y

\displaystyle A'=\frac{\frac{1}{2}-6y}{4}

Equate to 0

\displaystyle \frac{\frac{1}{2}-6y}{4}=0

Solving for y

\displaystyle y=\frac{1}{12}

And also

\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}

Compute the second derivative

\displaystyle A''=-\frac{3}{2}.

Since it's always negative, the point is a maximum

Thus, the maximum area is

\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}

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3 years ago
What is the answer for 14x divided by 7x
Neporo4naja [7]

Answer:

2x

Step-by-step explanation:

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Will mark brainliest
Anna007 [38]

Answer: x\leq 3

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Step-by-step explanation:

1) Simplify both sides of the inequality:

4x+8\geq 5x+5

2) Subtract 5x from both sides:

4x+8-5x\geq 5x+5-5x

3) Simplify:

-x+8\geq 5

4) Subtract 8 from both sides:

-x+8-8\geq 5-8

5) Simplify:

-x\geq -3

6) Divide both sides by -1:

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In your biology class, your final grade is based on several things: a lab score, scores on two major tests, and your score on th
Serggg [28]

Answer:

The weighted average is of 69.94.

Step-by-step explanation:

Weighted average:

The weighed average is found multiplying each grade by its respective weight.

The grades, and weights are:

67 on the lab, with a weight of 23% = 0.23

69 on the first major test, with a weight is 21.5% = 0.215

85 on the second major test, with a weight is 21.5% = 0.215.

63 on the final exam, with a weight of 34% = 0.34.

Weighted average:

A = 0.23*67 + 0.215*69 + 0.215*85 + 0.34*63 = 69.94

The weighted average is of 69.94.

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