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ikadub [295]
3 years ago
14

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other

Mathematics
2 answers:
Sonbull [250]3 years ago
4 0

Answer:

The lengths are

8 + 8[root]3 , 12 + 8[root]3 and 20 + 8[root]3.

Step-by-step explanation:

Let the side against the barn be xft

We were told that the adjacent side is 4ft more than the side against the barn. This means it has a length of (x + 4)ft.

Lastly, we are told that the hypotenuse is 8ft longer than the side along the barn I.e (x + 4 + 8)ft = ( x + 12)ft.

Since the three lengths form a triangle, the square of the hypotenuse equals the square of the adjacent side plus the square of the opposite side. This is the pythagoras' theorem.

Mathematically, this can be written as follows:

(x + 12)^2 = x^2 + (x + 4)^2

This yields x^2 + 24x + 144 = x^2 + x^2 + 8x + 16

x^2 + 24x + 144 = 2x^2 + 8x + 16

Rearranging this will yield:

x^2 - 16x - 128 = 0

Using the quadratic formula yields the following values for x :

8 - 8[root]3 and 8 + 8[root 3]

Since the first is negative, we discard it and the only correct value of x is 8 + 8[root]3

Now the other length is 12 + 8[root]3 and 20 + 8[root]3.

eimsori [14]3 years ago
4 0

Answer:

20,16,12

Step-by-step explanation:

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Answer:

4.4

Step-by-step explanation:

16.8 was for 4 hour

divide 4 by 16.8

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Find cot θ if csc θ = negative square root of thirty seven divided by six and tan θ > 0.
Ket [755]
You can use the trigonometric identity  1+\cot^2{x}=\csc^2{x}.

1+\cot^2{\theta}=(\frac{-\sqrt{37}}{6} )^2 \\ 1+\cot^2{\theta}=\frac{37}{36} \\ \cot^2{\theta}=\frac{1}{36} \\
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1. Kalena was asked to prove ifx(x - 1)(x + 1) = x3 - X represents a polynomial identity. She
Verdich [7]

Answer:

C. Kalena made a mistake in Step 3. The justification should state: -x²

+ x²

Step-by-step explanation:

Given the function x(x - 1)(x + 1) = x3 - X

To justify kelena proof

We will need to show if the two equations are equal.

Starting from the RHS with function x³-x

First we will factor out the common factor which is 'x' to have;

x(x²-1)

Factorising x²-1 using the difference of two square will give;

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Note that for two real number a and b, the expansion of a²-b² using difference vof two square will give;

a²-b² = (a+b)(a-b) hence;

Factorising x²-1 using the difference of two square will give;

x(x+1)(x-1)

Factorising x(x+1) gives x²+x, therefore

x(x+1)(x-1) = (x²+x)(x-1)

(x²+x)(x-1) = x³-x²+x²-x

The function x³-x²+x²-x gotten shows that kelena made a mistake in step 3, the justification should be -x²+x² not -x-x²

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