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)
Step-by-step explanation:
first, add the whole numbers
6 + 7 = 13
then the fractions
3/8 + 5/6
we need to get common denominators
do you know how to do that
2 71
3 86
4 101
6 131
.................
Answer:
12x - 6 ≤ - 18
Step-by-step explanation:
We will call "a number" x so
Twelve times a number would be:
12x
Then minus 6
12x - 6
Now we add the less than or equal to sign.
12x - 6 ≤
This is equal to negative eighteen, so:
12x - 6 ≤ - 18
As a bonus we can also solve for x.
12x - 6 = - 18 Add 6 to each side.
12x - 6 + 6 = -18 + 6
12x = -18 + 6
12x = -12 Divide each side by 12
12x/12 = -12/12
x = -12/12
x = -1
So we can say when x ≤ -1 Then 12x - 6 ≤ -18
Answer:
x=0.6241
Step-by-step explanation:
Square both sides and end up with
75.24+x=75.8641
isolate x
x=75.8641 - 75.24
x=0.6241
