The value (y2) of r = 1/2.
This question relates to equation of a straight line but we are looking for the slope in this case.
<h3>Slope</h3>
This is the ratio to which we measure the change along the y-axis to the change along the x-axis. The formula is given as

Data given;
- x1 = 3
- x2 = -3
- y1 = 5
- y2 = r
- m = 34 or 3/4
Substitute the values into the equation and solve for the unknown
NB; we are assuming that the slope here is 34 and not 3/4.

cross multiply both sides and make r the subject of formula

the value of r = -199.
This is not mathematically obtainable and we would use 3/4 as the value of slope

The most logical value of r = 1/2
Learn more on slope here;
brainly.com/question/4074386
Hmmm...
We have to divide 12 by 8, since we need to see how 8 students will share 12 slices.
12/8=1 and 1/2
So, each student can have a slice and a half, but since we aren't allowed to cut the slices...
We can let each student have one slice, with 4 left over, so...
Every student can have one slice, and then half can have a second slice :)!
I'm not sure if that is what was meant by the question, but that was all I could think of.
I hope I helped! :)
Answer:
f ) x+72=90(complementary angle)
x= 90-72
x=18
g )x+77=90
x=90-77
x=13
plz give me thanks
Answer:

Step-by-step explanation:
we would like to figure out the differential coefficient of 
remember that,
the differential coefficient of a function y is what is now called its derivative y', therefore let,

to do so distribute:

take derivative in both sides which yields:

by sum derivation rule we acquire:

Part-A: differentiating $e^{2x}$

the rule of composite function derivation is given by:

so let g(x) [2x] be u and transform it:

differentiate:

substitute back:

Part-B: differentiating ln(x)•e^2x
Product rule of differentiating is given by:

let
substitute

differentiate:

Final part:
substitute what we got:

and we're done!