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sp2606 [1]
3 years ago
5

4. Bryce has 220 feet of fencing that will enclose a rectangular corral. One side of the corral will be 48 ft long. What will be

the area of the corral?
Mathematics
2 answers:
RideAnS [48]3 years ago
7 0

Answer:

The area of the rectangular coral = 2,976 ft²

Step-by-step explanation:

Bryce has 220 ft of fencing to fence a rectangular coral.

Let the dimensions of the corral be x ft. × y ft.

One side of the coral is 48 ft. long

A rectangle has 4 sides, with each of the two opposite sides with the same dimension. Hence, the perimeter of the rectangular coral = 2(x + y) = 2x + 2y.

Total length of material for fencing = 220 ft.

Hence the perimeter of the reef = 220 ft.

2x + 2y = 220

And one length of the rectangular coral = x = 48 ft.

We can solve for the remaining dimension of the rectangular coral this way.

2(48) + 2y = 220

2y = 220 - 96 = 124

y = (124/2) = 62 ft.

Hence, the area of the rectangular coral = xy = 48 × 62 = 2,976 ft²

Hope this Helps!!!

allochka39001 [22]3 years ago
6 0

Answer:

2,976 ft2

Step-by-step explanation:

The perimeter of the corral needs to be 220 feet, so:

2*length + 2*width = 220 feet

length + width = 110 feet

If one side of the corral will be 48 feet (let's say the width), we have that:

length + 48 = 110

length = 110 - 48 = 62 feet

So the area of the corral is:

Area = length * width = 62 * 48 = 2,976 ft2

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equation: y=35+7(5)

Step-by-step explanation:

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Which two numbers have 50.35 as thier absolute value​
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In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t 2 , 2 + t − 5t 2 , 1 + 2t} to the standard basis of P2.
Serjik [45]

Complete Question:

In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t² , 2+t− 5t² , 1 + 2t} to the standard basis C = {1, t, t²}. Then, write t² as a linear combination of the polynomials in B.

Answer:

The change of coordinate matrix is :

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

Step-by-step explanation:

Let U =  {D, E, F} be any vector with respect to Basis B

U = D [1 − 3t²] + E [2+t− 5t²] + F[1 + 2t]..............(*)

U = [D+2E+F]+ t[E+2F] + t²[-3D-5E]...................(**)

In Matrix form;

\left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right] \left[\begin{array}{ccc}D\\E\\F\end{array}\right] = \left[\begin{array}{ccc}D+2E+F\\E+2F\\-3D-5E\end{array}\right]

The change of coordinate matrix is therefore,

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

To find D, E, F in (**) such that U = t²

D + 2E + F = 0.................(1)

E + 2F = 0.........................(2)

-3D -5E = 1........................(3)

Substituting eqn (2) into eqn (1 )

D=3F...................................(4)

Substituting equations (2) and (4) into eqn (3)

-9F+10F=1

F = 1

Put the value of F into equations (2) and (4)

E = -2(1) = -2

D = 3(1) = 3

Substituting the values of D, E, and F into (*)

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

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Answer:

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Step-by-step explanation:

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