Alan

As follows:

"You cannot solve the equation for F because it has a fraction in it." ???!!!!

F = 32 + 9*C/5   put Celsius in on the right hand side and get Fahrenheit out on the left hand side.

Here's a matrix solution:

Here is |v + w|:

Now you try the others using the same method.

the measure of one angle of a triangle is three times the measure of a second angle. the measure of the third angle is 12 less than the sum of the other two. what are the measures of the 3 angles?

Let the angles be A, B and C

We have three equations connecting them:

A+B+C = 180

A = 3B

C = A+B-12

(I'm assuming the 12 is degrees not radians).

Can you take it from here?

Write the equation of the line perpendicular to 2x-3y=12 that passes through the point (16,-7)

Rewrite the equation as  y = (2/3)x - 4  This has slope 2/3

A perpendicular line will have slope -3/2 so the equation of a perpendicuar line is y = -(3/2)x + c where c is a constant found from the fact that the line must go through point (16, -7).

-7 = -(3/2).16 + c

c = 24 - 7 

c = 17

Hence perpendicular line has equation  y = -(3/2)x + 17

Because triangle ABC is isosceles, angle ABC = angle ACB = (180 - 30)/2 = 75°

Angles on a straight line must sum to 180°, so 4x + 2x + 75 = 180  giving x = 17.5°, which makes angle ABE = 35°.  This doesn't match any of your options!

Let z = (sqrt(6) - sqrt(2)) + (sqrt(6) + sqrt(2))*i.  Compute z^(24).

Convert to polar notation 

z = r.exp(i.θ)   where r = √(√6 - √2)² + (√6 + √2)² ) = 4  and  θ = tan^-1( (√6 + √2)/ (√6 - √2) ) = 75°

z²⁴ = r²⁴.exp(i.24.θ) = 4²⁴.exp(1800°) = 4²⁴.(cos(1800°) + i.sin(1800°)) = 4²⁴.(cos(0°) + i.sin(0°)) = 4²⁴  

Let the lower number be n. Then

5*(n + n+ 2) = n*(n + 2) - 10      10n + 10 = n^2 +2n -10

Rearrange  

n^2 - 8n - 20 = 0

This factors as (n + 2)*(n - 10) = 0

I'll let you take it from here.

It isn't a minimum of the original cubic function, it is the point of steepest negative slope.  It is where the changing slope has a minimum. Try plotting the function for a range of x values, and then plot the slope over the same range of x values ( say, from x = -2 to 2)  to see.