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CaHeK987 [17]
3 years ago
12

Samuel plans to install a fence around the perimeter of his yard. His yard is shaped like a square and has an area of 40,000 squ

are feet. The company that he hires charges $2.50 pe foot for the fencing and $50.00 for the installation fee. What will be the cost of the fence, , in dollars?
Mathematics
1 answer:
NISA [10]3 years ago
7 0

Answer:

$450

Step-by-step explanation:

The area of the square is 40,000 square feet so the square's side length is 20 (√40,000).

The company charges $2.50 per foot, and the perimeter is 80 (20*4), so the  cost is $400 ($2.50 * 8).

Add on the $50 installation fee and the total cost is $450 ($400 + $50).

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Answer:  the slope is 2/1 or 2.

Step-by-step explanation:

The slope is the difference in the y coordinates divided by the difference in the x coordinates so you have to divide 6 by 3 since they have already given you the difference.

6/3 = 2

4 0
3 years ago
Simplify the expression please!<br><br> -5+i/2i
melamori03 [73]

Answer:-9/2 or -4.

5

Step-by-step explanation:

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3 years ago
What is the surface area of this design?<br> A.150in2<br> B.197in2<br> C.210in2<br> D.127in2
ANEK [815]

Answer:

197 in ^2 (answer B of the list)

Step-by-step explanation:

Notice that this figure has a total of 6 faces, four of which are rectangles (whose area is calculated as "base times height") and two trapezoids (whose area is (B+b)H/2 ).

The total surface area is therefore the addition of these six areas:

Rectangles:

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5 in x 5 in = 25 in^2

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Trapezoids:

Two of equal dimensions: B = 9 in, b = 5 in, H = 5 in

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6 0
3 years ago
Read 2 more answers
Obtain the general solution to the equation. (x^2+10) + xy = 4x=0 The general solution is y(x) = ignoring lost solutions, if any
alukav5142 [94]

Answer:

y(x)=4+\frac{C}{\sqrt{x^2+10}}

Step-by-step explanation:

We are given that a differential equation

(x^2+10)y'+xy-4x=0

We have to find the general solution of given differential equation

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Compare with

y'+P(x) y=Q(x)

We get

P(x)=\frac{x}{x^2+10}

Q(x)=\frac{4x}{x^2+10}

I.F=e^{\int\frac{x}{x^2+10} dx}=e^{\frac{1}{2}ln(x^2+10)}

e^{ln\sqrt(x^2+10)}=\sqrt{x^2+10}

y\cdot \sqrt{x^2+10}=\int \frac{4x}{x^2+10}\times \sqrt{x^2+10} dx+C

y\cdot \sqrt{x^2+10}=\int \frac{4x}{\sqrt{x^2+10}}+C

y\cdot \sqrt{x^2+10}=4\sqrt{x^2+10}+C

y(x)=4+\frac{C}{\sqrt{x^2+10}}

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