Answer:
the two positive consecutive integers are 4 and 6.
Step-by-step explanation:
Let the smaller integer be s; then s^2 = (s + 2) + 10.
Simplifying, s^2 - s - 2 - 10 = 0, or
s^2 - s - 12 = 0.
Solve this by factoring: (s - 4)(s + 3) = 0.
Then s = 4 and s = -3.
If the first even integer is 4, the next is 6. We omit s = -3 because it's not even.
The smaller integer is 4. Does this satisfy the equation s^2 = (s + 2) + 10?
4^2 = (4 + 2) + 10 True or False?
16 = 6 + 10 = 16.
True.
So the two positive consecutive integers are 4 and 6.
Vertical angles are congruent, so
107=3x-4
3x=111
x=111/3=37°
I don’t know I just want points bye hdhdjjdjddjdjdjjjfjtfjgktk
58 I just took the test and got it right
![\bf \stackrel{\textit{multiplying both sides by LCD of 3}}{3(y+5)=3\left[ \cfrac{5}{3}(x-3) \right]}\implies 3y+15=5(x-3) \\\\\\ 3y+15=5x-15\implies -5x+3y=-30\implies \stackrel{\textit{multiplying by -1}}{5x-3y=30}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20LCD%20of%203%7D%7D%7B3%28y%2B5%29%3D3%5Cleft%5B%20%5Ccfrac%7B5%7D%7B3%7D%28x-3%29%20%5Cright%5D%7D%5Cimplies%203y%2B15%3D5%28x-3%29%0A%5C%5C%5C%5C%5C%5C%0A3y%2B15%3D5x-15%5Cimplies%20-5x%2B3y%3D-30%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20by%20-1%7D%7D%7B5x-3y%3D30%7D)
bearing in mind the standard form uses all integers, and the x-variable cannot have a negative coefficient.
