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2 years ago

Select the correct answer. What is true of the function as the x values increase?

Mathematics

2 answers:

Alinara [238K]2 years ago

5 0

Answer:

Step-by-step explanation:

Brums [2.3K]2 years ago

4 0

Answer:

Step-by-step explanation:

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Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8

Gekata [30.6K]

First, let $(x,y)$ be a point in our parabola. Since we know that the focus of our parabola is the point (0,8), we are going to use the distance formula to find the distance between the two points:
$\sqrt{(x-0)^{2}+(y-8)^{2}}$

Next, we are going to find the distance between the directrix and the point in our parabola. Remember that the distance between a point (x,y) of a parabola and its directrix, $y=c$ , is: $|y-c|$ . Since our directrix is y=-8, the distance to our point will be:
$|y-(-8)|$
$|y+8|$

Now, we are going to equate those two distances, and square them to get rid of the square root and the absolute value:
$\sqrt{(x-0)^{2}+(y-8)^{2}}=|y+8|$
$(\sqrt{(x-0)^{2}+(y-8)^{2}})^{2}=|y+8|^{2}$
$(x-0)^{2}+(y-8)^{2}=y+8$

Finally, we can expand and solve for $y$ :
$x^2+y^2-16y+64=y^2+16y+64$
$x^{2}-16y=16y$
$32y=x^{2}$
$y= \frac{1}{32} x^2$

We can conclude that t<span>he standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8 is </span> $y= \frac{1}{32} x^{2}$

3 0

3 years ago

A company packs 36 boxes in each case to ship to stores. You have been assigned the task to determine the least amount of volume

pshichka [43]

Answer:

432 in^3.

Step-by-step explanation:

The volume of each individual box is 12 cubic inches. 12 × 36 = 432 cubic inches. The case needs to hold at least 432 in^3.

4 0

3 years ago

I need help with this, pls help

Alex

Answer:

Step-by-step explanation:

why

4 0

2 years ago

Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,

Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j} = {(1,0), (0,1)}

is a base in $\mathbb{R}^2$

and F is linear, then

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0

3 years ago

Please help, see attatched file.

liubo4ka [24]

The answer above is correct.

4 0

3 years ago