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Ymorist [56]
4 years ago
5

What is the value of x?

Mathematics
1 answer:
Archy [21]4 years ago
7 0

Answer:

<h2>x=12</h2>

solution,

We have ABCD~FECG

Sides of similar shapes in proportion.

Here,

Setting proportion of sides of similar shapes

\frac{ab}{fe}  =  \frac{bc}{ec}

\frac{x}{6}   =  \frac{10}{5} (substituting \: value) \\ x \times 5 = 10 \times 6(cross \: multiplication) \\ or \: 5x = 60 \\ or \: x =  \frac{60}{5}  \\ x = 12

Hope this helps...

Good luck on your assignment..

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Exactly one solution

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A certain forest can support a population of 800 deer. There are currently 200 deer in the forest and their population is growin
Otrada [13]

It would take 70 years for the deer population to reach 800.

An exponential function is in the form:

y = abˣ

where b is the multiplier, y, x are variables and a is the initial value.

Let y represent the population and x represent the time in years.

a = 200, b = 100% + 2% = 102% = 1.02

y = 200(1.02)ˣ

For a population of 800:

800 = 200(1.02)ˣ

ln(4) = xln(1.02)

x = 70 years

It would take 70 years for the deer population to reach 800.

Find out more on exponential function at: brainly.com/question/2456547

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to convert 135 minutes into hours, which ratio could you multiply by? Remember the 1 hour = 60 minutes
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what is the amount returned on a deposit of $1,000 held for 2 years at 10% interest compound annually? ​
pickupchik [31]

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3 years ago
Let the (x; y) coordinates represent locations on the ground. The height h of
grigory [225]

The critical points of <em>h(x,y)</em> occur wherever its partial derivatives h_x and h_y vanish simultaneously. We have

h_x = 8-4y-8x = 0 \implies y=2-2x \\\\ h_y = 10-4x-12y^2 = 0 \implies 2x+6y^2=5

Substitute <em>y</em> in the second equation and solve for <em>x</em>, then for <em>y</em> :

2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}

This is to say there are two critical points,

(x,y)=\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\text{ and }(x,y)=\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)

To classify these critical points, we carry out the second partial derivative test. <em>h(x,y)</em> has Hessian

H(x,y) = \begin{bmatrix}h_{xx}&h_{xy}\\h_{yx}&h_{yy}\end{bmatrix} = \begin{bmatrix}-8&-4\\-4&-24y\end{bmatrix}

whose determinant is 192y-16. Now,

• if the Hessian determinant is negative at a given critical point, then you have a saddle point

• if both the determinant and h_{xx} are positive at the point, then it's a local minimum

• if the determinant is positive and h_{xx} is negative, then it's a local maximum

• otherwise the test fails

We have

\det\left(H\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\right) = -16\sqrt{73} < 0

while

\det\left(H\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)\right) = 16\sqrt{73}>0 \\\\ \text{ and } \\\\ h_{xx}\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-8 < 0

So, we end up with

h\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-\dfrac{4247+37\sqrt{73}}{72} \text{ (saddle point)}\\\\\text{ and }\\\\h\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)=-\dfrac{4247-37\sqrt{73}}{72} \text{ (local max)}

7 0
3 years ago
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