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4 years ago

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]4 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]4 years ago

4 0

Answer:

20,16,12

Step-by-step explanation:

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Free_Kalibri [48]

Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).

<h3>What are the mean and the standard error for each sample?</h3>

Considering the data given:

$\mu_S = 3.8, n = 175, s_S = \frac{1.7}{\sqrt{175}} = 0.1285$

$\mu_N = 4.8, n = 152, s_N = \frac{1.2}{\sqrt{152}} = 0.0973$

<h3>What is the mean and the standard error for the distribution of differences?</h3>

The mean is the subtraction of the means, hence:

$\overline{x} = \mu_S - \mu_N = 3.8 - 4.8 = -1$

The standard error is the square root of the sum of the variances of each sample, hence:

$s = \sqrt{s_S^2 + s_N^2} = \sqrt{0.1285^2 + 0.0973^2} = 0.1612$

<h3>What is the confidence interval?</h3>

It is given by:

$\overline{x} \pm zs$

We have a 95% confidence interval, hence the critical value is of z = 1.96.

Then, the bounds of the interval are given as follows:

$\overline{x} - zs = -1 - 1.96(0.1612) = -1.3$

$\overline{x} + zs = -1 + 1.96(0.1612) = -0.7$

More can be learned about the z-distribution at brainly.com/question/25890103

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2 years ago

Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$ <span />

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3 years ago