Let the slower runners speed be X kilometers per hour.
Then the faster runners speed would be X+2 kilometers per hour.
The formula for distance is Speed times time.
The distance is given as 30 kilometers and time is given as 3 hours.
Since there are two runners you need to add the both of them together.
The equation becomes 30 = 3x + 3(x+2)
Now solve for x:
30 = 3x + 3(x+2)
Simplify:
30 = 3x + 3x +6
30 = 6x + 6
Subtract 6 from each side:
24 = 6x
Divide both sides by 6:
x = 24/6
x = 4
The slower runner ran at 4 kilometers per hour.
The faster runner ran at 4+2 = 6 kilometers per hour.
Answer:

Step-by-step explanation:
To start, lets form our polynomial factors from our zeroes.
x = -2
x + 2 = 0
x = 5
x - 5 = 0
Now lets put these into factored form by moving them into parenthesis and multiplying them with each other!
(x+2)(x-5)
We can start moving them out! I am going to use FOIL, but you could also use distributive or any other method/property.
x · x = x²
x · -5 = -5x
x · 2 = 2x
2 · 5 = 10
Now lets put our terms into our polynomial function, based on the order of the powers.

We are done! Hope this helps!
-29/30 +7/12÷1/2
7/12÷1/2=7/6
-29/30 + 7/6=1/5
1/5
Check the picture below.
let's notice that the angle at K is an inscribed angle with an intercepted arc
![\bf \stackrel{\textit{using the inscribed angle theorem}}{K=\cfrac{\widehat{LI}+\widehat{IJ}}{2}}\implies 9x+1=\cfrac{(10x-1)+59}{2} \\\\\\ 9x+1=\cfrac{10x+58}{2}\implies 18x+2=10x+58\implies 8x+2=58 \\\\\\ 8x=56\implies x=\cfrac{56}{8}\implies x=7 \\\\[-0.35em] ~\dotfill\\\\ K=9x+1\implies K=9(7)+1\implies \boxed{K=64}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20inscribed%20angle%20theorem%7D%7D%7BK%3D%5Ccfrac%7B%5Cwidehat%7BLI%7D%2B%5Cwidehat%7BIJ%7D%7D%7B2%7D%7D%5Cimplies%209x%2B1%3D%5Ccfrac%7B%2810x-1%29%2B59%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%209x%2B1%3D%5Ccfrac%7B10x%2B58%7D%7B2%7D%5Cimplies%2018x%2B2%3D10x%2B58%5Cimplies%208x%2B2%3D58%20%5C%5C%5C%5C%5C%5C%208x%3D56%5Cimplies%20x%3D%5Ccfrac%7B56%7D%7B8%7D%5Cimplies%20x%3D7%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20K%3D9x%2B1%5Cimplies%20K%3D9%287%29%2B1%5Cimplies%20%5Cboxed%7BK%3D64%7D)
now, let's notice something again, the angle at L is also an inscribed angle, intercepting and arc of 97 + 59 = 156, so then, by the inscribed angle theorem,
∡L is half that, or 78°.
now, let's take a look at the picture down below, to the inscribed quadrilateral conjecture, since ∡J and ∡I are both supplementary angles, then
∡I = 180 - 64 = 116°.
∡J = 180 - 78 = 102°.
Answer:
Find the exact value using trigonometric identities.
100
Step-by-step explanation: