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Genrish500 [490]
3 years ago
9

X>= -5 and y>= -3 draw a graph with the following domain and range restrictions.

Mathematics
1 answer:
IRISSAK [1]3 years ago
3 0

Answer:(-5,-3) -5 to left -3 down

Step-by-step explanation:

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Please see attached below. <br> Please explain.
vivado [14]

Answer:

15 rentals

Step-by-step explanation:

You can (and may be expected to) set up an equation that equates the total cost at one store to the total cost at the other store. When you work through the solution of this equation, you find that the "break even" number of rentals is the ratio of the difference in fixed cost (setup fee) to the difference in per-use cost (rental charge).

Here, that ratio is ...

(15.00 -7.50)/(2.25 -1.75) = 7.50/0.50 = 15

15 rentals will make the total costs the same.

5 0
3 years ago
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Can someone help me with this problem please? I'll mark you as brainliest!
leonid [27]

Answer:

put a closed for on -1 and go to the left with an arrow

3 0
3 years ago
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Rebecca went on a 240 mile trip to a soccer game. On the way back, due to road construction she had to drive 12 miles per hour s
AlladinOne [14]

Answer:

20mph

Step-by-step explanation:

because you have to divide 240 by 12 and the answer is 20

8 0
3 years ago
A company manufactures running shoes and basketball shoes. The total revenue (in thousands of dollars) from x1 units of running
Alborosie

Answer:

x_1 =2 , x_2=7

Step-by-step explanation:

Consider the revenue function given by R(x_1,x_2) = -5x_1^2-8x_2^2 -2x_1x_2+34x_1+116x_2. We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).

\frac{dR}{dx_1} = -10x_1-2x_2+34 =0

\frac{dR}{dx_2} = -16x_2-2x_1+116 =0

From the first equation, we get, x_2 = \frac{-10x_1+34}{2}.If we replace that in the second equation, we get

-16\frac{-10x_1+34}{2} -2x_1+116=0= 80x_1-2x_1+116-272= 78x_1-156

From where we get that x_1 = \frac{156}{78}=2. If we replace that in the first equation, we get

x_2 = \frac{-10\cdot 2 +34}{2}=\frac{14}{2} = 7

So, the critical point is (x_1,x_2) = (2,7). We must check that it is a maximum. To do so, we will use the Hessian criteria. To do so, we must calculate the second derivatives and the crossed derivatives  and check if the criteria is fulfilled in order for it to be a maximum. We get that

\frac{d^2R}{dx_1dx_2}= -2 = \frac{d^2R}{dx_2dx_1}

\frac{d^2R}{dx_{1}^2}=-10, \frac{d^2R}{dx_{2}^2}=-16

We have the following matrix,  

\left[\begin{matrix} -10 & -2 \\ -2 & -16\end{matrix}\right].

Recall that the Hessian criteria says that, for the point to be a maximum, the determinant of the whole matrix should be positive and the element of the matrix that is in the upper left corner should be negative. Note that the determinant of the matrix is (-10)\cdot (-16) - (-2)(-2) = 156>0 and that -10<0. Hence, the criteria is fulfilled and the critical point is a maximum

8 0
3 years ago
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Simplify each expression justify each step 3x•(7•x)
jarptica [38.1K]
Jdjdbdjdbdbbssnskskbsdnenen
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3 years ago
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