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3 years ago

Please help thank you.

Mathematics

1 answer:

miv72 [106K]3 years ago

4 0

Hello!

All lines within a circle are the same length

We first have to find the length of one of the lines.

7 + 4 = 11

Now we find what we are missing from the second line

5 + z = 11

z = 6

The answer is C)6

Hope this helps!

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Solve the inequality. Graph the solution +5<2 The solution is a 6

BartSMP [9]

Answer:

x<-3

The graph in the attached figure

Step-by-step explanation:

The correct question is

Solve the inequality. Graph the solution.

x+5<2

we have

$x+5$

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Subtract 5 both sides

$x$

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All real numbers less than -3

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3 years ago

Cheryl has twice as many nickels as quarters and five more dimes than nickels.

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1.60

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3 years ago

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OleMash [197]

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3 years ago

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Hat is the equation of the line through the point (5,-6) slope of-3/5

lutik1710 [3]

Answer:y=-3/5X-9

Step-by-step explanation:

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3 years ago

Find the absolute maximum and minimum values of the following function on the specified region R.

yan [13]

Answer:

Step-by-step explanation:

Since it is said that the region R is a semicircular disc, we asume that the boundaries of the region are given by $-2\leq x \leq 2, 0\leq y \leq \sqrt[]{4-x^2}$ . First, we must solve the optimization problem without any restrictions, and see if the points we get lay inside the region of interest. To do so, consider the given function F(x,y). We want to find the point for which it's gradient is equal to zero, that is

$\frac{dF}{dx}=9y=0$

$\frac{dF}{dy}=9x=0$

This implies that (x,y) = (0,0). This point lays inside the region R. We will use the Hessian criteria to check if its a minimum o r a maximum. To do so, we calculate the matrix of second derivates

$\frac{d^2F}{dx^2} = 0 = \frac{d^2F}{dy^2}$

$\frac{d^2F}{dxdy} = 9 = \frac{d^2F}{dydx}$

so we get the matrix

$\left[\begin{matrix} 0 & 9 \\ 9 & 0\end{matrix}\right]$

Note that the first determinant is 0, and the second determinant is -9. THis tell us that the point is a saddle point, hence not a minimum nor maximum.

Since the function is continous and the region R is closed and bound (hence compact) the maximum and minimum must be attained on the boundaries of R. REcall that when $-2\leq x \leq x$ and y=0 we have that F(x,0) = 0. So, we want to pay attention to the critical values over the circle, restricting that the values of y must be positive. To do so, consider the following function

$H(x,y, \lambda) = 9xy - \lambda(x^2+y^2-4)$ which consists of the original function and a function that describes the restriction (the circle x^2+y^2=4), we want that the gradient of H is 0.

Then,

$\frac{dH}{dx} = 9y-2\lambda x =0$

$\frac{dH}{dx} = 9x-2\lambda y =0$

$\frac{dH}{d\lambda} = x^2+y^2-4 =0$

From the first and second equation we get that

$\lambda = \frac{9y}{2x} = \frac{9x}{2y}$

which implies that $y^2=x^2$ . If we replace this in the restriction, we have that $x^2+x^2 = 2x^2 = 4$ which gives us that $x=\pm \sqrt[]{2}$ . Since we only care for the positive values of y, and that $y=\pm x$ , we have the following critical points $(\sqrt[]{2},\sqrt[]{2}), (-(\sqrt[]{2},\sqrt[]{2})$ . Note that for the first point, the value of the function is