CPhill

x^2 + 2y^2 = 16   solve for y..... subtract x^2 from both sides

2y^2 = 16 - x^2     divide both sides by 2

y^2 = 8 -x^2/2      take the positive root of both sides

y = (8 - x^2/2)^(1/2)

Now  we wish to maximize xy....so we have

x(8 - x^2/2)^(1/2)    take the derivative and set this to 0

(8 - x^2/2)^(1/2) -(1/2)(x^2)(8 - x^2/2)^(-1/2) = 0   factor

(8- x^2)^(-1/2)[(8 - x^2/2) -(1/2)(x^2)] = 0

So we have

[(8 - x^2/2) -(1/2)(x^2)] = 0

8 - x^2 = 0

x^2 = 8

x =√8  = 2√2

So y = (8 - 8/2)^(1/2) = (8 - 4)^(1/2) = (4)^(1/2) = 2

And the max product =   (2√2)(2) = 4√2 = about 5.65685

You have the slope a little incorrect ...we have   (-13 - -1) /(-6-6) = (-13 +1) /(-12)  = -12/-12  = 1....whew!!! that makes things easier !!!  so we have.......

y - (-13) = (1) ( x - (-6))    simplify

y + 13 = x + 6        subtract 13 from both sides

y = x + 6 - 13      simplify

y = x - 7     that's slope-intercept form   .........add 7 to both sides and subtract y from both sides

7 = x - y  →    x - y = 7     and that's standard form

Hope that helps.....BTW.....you can "plug" in the points to make sure the equation is correct....I think you will find it is....

Solve ln (2x + 3) = 7   in exponential form, we have

e^7 = 2x + 3       subtract 3 from both sides

e^7 - 3  = 2x       divide by 2 on both sides

[e^7 - 3] / 2  = x   = about 546.817

Just set the onsite calculator to radian mode at the bottom left...... key in tan(4) and hit "="

Well....I don't know what we did......but we're always glad to hear success stories !!!!

I assume you have  (1/4)x^3...if so......forget the 1/4 part for a second

And the derivative of x^3, by the Power Rule, is just 3x^2....so...multiplying this by the 1/4, we have

(1/4)*3x^2  =   (3/4)x^2       and that's it.....

Here's the last one

lim x→ 0   [3x - tan^2(7x)] / (2x)      you have the first part correct........let's look at this part

lim x→ 0   [ - tan^2(7x)] / (2x)     note we can write this as.....

lim x→ 0    -[ tan(7x)] * [tan(7x)] / (2x)...note, we can use a "trick" here for the second expression.....multiply the top and bottom by 7....this gives

lim x→ 0    -[ tan(7x)] * [7tan(7x)] / (7*2x)  =  lim x→ 0   -[ tan(7x)] * [7tan(7x)] / (2 * 7x) =

lim x→ 0   (-7/2)[ tan(7x)] * [tan(7x)] / ( 7x)      and concentrating on  [tan(7x)] / ( 7x), we can write

sin(7x)/[cos(7x)*7x] = sin(7x)/7x *1/ cos(7x)]   and using an identity,  lim x → 0   sin(7x)/7x) = 1, and

lim x→ 0   1/ cos(7x)  = 1...so we're left with

lim x→ 0   (-7/2)[ tan(7x)]   and as x → 0, tan(7x) → 0    so.... (-7/2)* 0 = 0

And your final answer of 3/2 is correct.....!!!!

Oops....I'll do an edit to the original post.....sorry!!!!

A line going through (a, b) with a slope of 0 has the equation y = b

So  a line going through (-5, -2) with a slope of 0 has the equation  y = -2

What is the number on "x" in the "tan^2"  term??