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2 years ago

Write an equation of a line that passes through (-1,6) and has a slope of 4.

Mathematics

1 answer:

jeka942 years ago

7 0

Answer:

y-6=4(x+1) or y=4x+10

Step-by-step explanation:

Point: (-1,6)

m=4

Slope-point form is y-y1=m(x-x1)

y1=6

x1=-1

Now, plug it in!

y-6=4(x-(-1)) which is y-6=4(x+1)

Of course, if you want you could always simplify it down to y=4x+10 if you wanted to

Hope this helps!

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A company produces two types of solar panels per year: x thousand of type A and y thousand of type B. The revenue and cost equat

jolli1 [7]

Answer:

7000 type A
6000 type B

Step-by-step explanation:

We assume your equations are intended to be ...

$C(x,y)=x^2-2xy+9y^2+3x-87y-6\\R(x,y)=5x+7y$

Then the profit equation is ...

$P(x,y)=R(x,y)-C(x,y)=5x+7y-x^2+2xy-9y^2-3x+87y+6\\=-x^2+2xy-9y^2+2x+94y+6$

The partial derivatives of profit with respect to x and y are zero when profit is maximized.

∂P/∂x = 0 = -2x +2y +2

∂P/∂y = 0 = 2x -18y +94

Simplifying, these equations are ...

y = x -1
x -9y = -47

Substituting the first into the second gives ...

x -9(x -1) = -47

-8x = -56

x = 7

y = 7 -1 = 6

The company will maximize profit by selling 7000 panels of type A and 6000 panels of type B.

5 0

3 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

For the first parenthesis:

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$ :

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

For the second parenthesis:

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

$k=4$

$r=\sqrt{100}=10$

8 0

3 years ago

3/7 times 0.1 divided by 5/21

Sati [7]

Answer:

0.000408163265306

Step-by-step explanation:

ITS SIMPLE JUST CACULATE IT

8 0

3 years ago

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Pre Calc Question Help,

julsineya [31]

Answer:

Step-by-step explanation:

Domain is the set of all values of x where the function is defined. For this function the only limitation upon x exists when the denominator would equal zero, because division by zero is undefined. So the denominator 13-x cannot equal zero thus x cannot equal 13. Thus the domain is all real numbers except 13.

5 0

3 years ago

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Question 1

matrenka [14]

I think it may be (10×-3)

7 0

3 years ago