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NISA [10]
3 years ago
8

Reflect the vector (-3,5) across the line y=x.

Mathematics
1 answer:
user100 [1]3 years ago
4 0

Answer:

(5, - 3 )

Step-by-step explanation:

Under a refection in the line y = x

a point x, y ) → (y, x ), then

(- 3, 5 ) → (5, - 3 )

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Option C. -3x^2-13x-5

Step-by-step explanation:

we know that

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so

(5x^{2}-7x-6)-(8x^{2}+6x-1)\\ \\= (5x^{2}-7x-6)-8x^{2}-6x+1\\ \\=(5x^{2}-8x^{2})+(-7x-6x)+(-6+1)\\ \\=-3x^{2} -13x-5

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The directions on a sewing pattern say to cut an extra 12% of fabric to account for error. Chole needs 0.8 yard of fabric to mak
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4 0
3 years ago
Read 2 more answers
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

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Answer:

q

Step-by-step explanation:

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3 years ago
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