CalculatorUser

2)

We can try to find the distance of the perihelion to the center of the ellipse.

116 / 2 = 58 million miles

58 - 46 (distance from perhelion to sun) = 12 million miles between sun and center of ellipse

Not sure about my answers never learned calculus

This is basically what the question is asking

Lets say the sun is located at the focus of the left side of the image

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So the distance between the perihelion and aphelion should be 46 + 70 = 116 million miles??

I haven't learned much in calculus yet this is as much as I can give you

Thanks melody!

Sorry blocked on my browser

HINT:

Notice how you can substitute in y= 4x - 19 into 2x + y = 95

 2x + 4x - 19 = 95.

Solve for x

I'm not sure I haven't taken algebra 2 yet, but I think its like htis

Since 2016 is even,

what is (1+i)^2?

What is (1-i)^2

So I think th answer should be 4i, after plugging that in the calc. But EP may be correct he probably took alg 2 already.

Is there any way you can post the image. When I copy pasted the link my browser marked it as illicit ****.    So I can't see it.

My browser blocked the image.

Where is K located? 

Do you have a picture? I mean, you don't need a pic, but where is K?

I will give you a hint,

whenever an equation is perendicular, it has the NEGATIVE RECIPROCAL of the original slope.

This is an example problem:

y = 2x + 9 and passes through ( 4, 2)?

So new slope is 2 =>   -1/2

y = -1/2x + b

2 = -4/2 + b             (Plug in coordinates (4, 2)

4 = b

So the equation is  y = -1/2x + 4

Don't simplify 10/75

First, do 1 - (10/75)

(75/75) - (10/75)   =    65/75

THEN simplify

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