To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
Answer: dimond
Step-by-step explanation:
Just add them as if it were normal then put the s behind it
Answer:
For #1, 8.9.
Step-by-step explanation:
I can attempt to answer it, but I can't promise anything.
Alright, so, the equation for this should be A²+B²=C²
C is usually the longer side, which for you is the side that is missing.
Let's try the first one. If what I see is right, the 2 sides are 4 and 8. A is 8 and B is 4.
That means that the equation would be 8²+4²=C²
That would be 64+16=80
The equation is now 80= c²
The opposite of "to the power of 2" is a square root. So, now you're going to find the square root of 80, which is 8.9 to the nearest tenth.
So, for the first question, the answer should be 8.9
I hope this helps you solve the rest.
