9We have such eqation
Q
/*9 (multiply both sides by 9)
9Q+5=
/-5 both sides
9Q=
9Q=
9Q=
9Q=-
/:9 divide both sides by 9
Q=-
Q![\frac{21}{6*9} =- \frac{21}{54}=[tex] \frac{7}{18}](https://tex.z-dn.net/?f=%20%5Cfrac%7B21%7D%7B6%2A9%7D%20%3D-%20%5Cfrac%7B21%7D%7B54%7D%3D%5Btex%5D%20%5Cfrac%7B7%7D%7B18%7D%20)
- its the answer
Q
9Q+5=
9Q=
9Q=
9Q=
9Q=-
Q=-
Q
What is the answer help
1 answer:
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Answer:
y=6x
Step-by-step explanation:
Let us assume the table is the one shown in the attachment.
We can observe the following pattern from the table:
Hence in general
Therefore the equation that could have been used to create this table is
The only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54
remember if we can do
xy=0 then assume x and y=0
so factor
0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is
if you have
ax^2+bx+c=0 then
x=
x^2+7x+18
a=1
b=7
c=18
x=
x=
x=
i=√-1
x=
the zerose would be
x=
or 
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54
remember if we can do
xy=0 then assume x and y=0
so factor
0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is
if you have
ax^2+bx+c=0 then
x=
x^2+7x+18
a=1
b=7
c=18
x=
x=
x=
i=√-1
x=
the zerose would be
x=