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Yuliya22 [10]
2 years ago
8

For his book report, Bobby made color copies and black-and-white copies. The black-and-white copies cost $2.75 in all. The color

copies cost $0.08 each. How many color copies did Bobby make if he spent $4.03 in all?
Mathematics
1 answer:
Feliz [49]2 years ago
3 0
16, i think it is the right answer.






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Consider the linear function that is represented by the equation y = 2 x + 2 and the linear function that is represented by the
Arisa [49]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
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Can someone please help ? :(
Anika [276]

Answer:

I'll list the answers as coordinates (just input the second number in each coordinate)

(-2, 11)

(0, 1)

(2, -9)

(4, -19)

Step-by-step explanation:

So you see how the x column is filled out? You plug in the numbers in each area and solve for y.

So we start off with -2. You take the x out and place -2 instead. Your equation should look like this:

y = -5 (-2) + 1

Next you multiply -5 and -2, giving you 10. Your equation would look like this now:

y = 10 + 1

Next you add 10 and 1 together, and you have your answer. Keep doing the same thing, but instead of -2 for x, use 0, 2, and 4.

Hope I helped! Have a nice day or night! ^-^

8 0
2 years ago
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Anyone help with this??
Brut [27]

Answer:

bottom 2 shapes

Step-by-step explanation:

8 0
3 years ago
Multiply the binomials (x + 1)(-5x + 10) *
Leona [35]
So i think here you would FOIL. so -5x^2+10x-5x+10 and then combine like terms :)
7 0
3 years ago
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Use lagrange multipliers to find the point on the plane x â 2y + 3z = 6 that is closest to the point (0, 2, 4).
Arisa [49]
The distance between a point (x,y,z) on the given plane and the point (0, 2, 4) is

\sqrt{f(x,y,z)}=\sqrt{x^2+(y-2)^2+(z-4)^2}

but since \sqrt{f(x,y,z)} and f(x,y,z) share critical points, we can instead consider the problem of optimizing f(x,y,z) subject to x-2y+3z=6.

The Lagrangian is

L(x,y,z,\lambda)=x^2+(y-2)^2+(z-4)^2+\lambda(x-2y+3z-6)

with partial derivatives (set equal to 0)

L_x=2x+\lambda=0\implies x=-\dfrac\lambda2
L_y=2(y-2)-2\lambda=0\implies y=2+\lambda
L_z=2(z-4)+3\lambda=0\implies z=4-\dfrac{3\lambda}2
L_\lambda=x-2y+3z-6=0\implies x-2y+3z=6

Solve for \lambda:

x-2y+3z=-\dfrac\lambda2-2(2+\lambda)+3\left(4-\dfrac{3\lambda}2\right)=6
\implies2=7\lambda\implies\lambda=\dfrac27

which gives the critical point

x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7

We can confirm that this is a minimum by checking the Hessian matrix of f(x,y,z):

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

\mathbf H is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.

At this point, we get a distance from (0, 2, 4) of

\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}
8 0
2 years ago
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