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DedPeter [7]
3 years ago
8

You have 80 yards of fencing to enclose a rectangular region find the dimensions of the rectangle that maximize the enclosed are

a
Mathematics
1 answer:
harkovskaia [24]3 years ago
3 0
A maximum area is always given by a square

so each side will be 80/4  = 20 yards long
You might be interested in
Y+4x=10x-6 solve for x
7nadin3 [17]

Answer:

(y-6)/6 = x

Step-by-step explanation:

y+4x=10x-6

Subtract 4x from each side

y+4x-4x=10x-4x-6

y = 6x-6

Add 6 to each side

y +6 = 6x-6+6

y+6 = 6x

Divide by 6

(y-6)/6 = 6x/6

(y-6)/6 = x



8 0
3 years ago
Read 2 more answers
290 divided by 70 estimate to
labwork [276]
290 divides by 70 is 4.14285...
Estimated to the ten thousandths: 4.1429
Estimated to the thousandths: 4.143
Estimated to the tenths: 4.14
4 0
3 years ago
Given the equation f(x) = (x + 3)2 – 8, fill in the blanks:
telo118 [61]
Axis of symmetry -3 and -8 or 3 and -8 

6 0
3 years ago
Pls help not to sure on this one
skad [1K]
The answer is A. U basically divide the price with the amount of item and pick the cheapest one :)
3 0
2 years ago
Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x
BaLLatris [955]

Answer:

The area of the region is 25,351 units^2.

Step-by-step explanation:

The Fundamental Theorem of Calculus:<em> if </em>f<em> is a continuous function on </em>[a,b]<em>, then</em>

                                   \int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) |  {_a^b}

where F is an antiderivative of f.

A function F is an antiderivative of the function f if

                                                    F^{'}(x)=f(x)

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x + 15 and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx

\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2

3 0
3 years ago
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