We have to add the equations in a clever way such that one of the variables cancels out. If you multiply the top one with 3 and the bottom one with 2, you cancel x. Like this:
3(2x+3y)=3*6 => 6x + 9y = 18
2(-3x+5y)=2*10 => -6x + 10y = 20
9y + 10y = 38 => 19y = 38 => y=2
If y=2 then 2x + 6 = 6 => x=0
The pair (0,2) is a solution.
Hello! And thank you for your question!
First we are going to redo the equation:
<span><span><span>y<span><span>^3</span><span></span></span></span>−3<span>y^<span><span>2</span><span></span></span></span>(3)+3(y)×<span>3<span><span>^2</span><span></span></span></span>−<span>3<span><span>^3</span><span></span></span></span>=0
Then we are going to use Cube of Difference:
0 = (y - 3)^3
That take cube root of both sides:
0 = y - 3
Finally add 3 to both sides:
3 = y
Final Answer:
y = 3</span></span>
Answer:
48.58
Step-by-step explanation:
I am on a time crunch so I am unable to explain right now. But if you really need help understanding, message me! :D
Answer:
16
Step-by-step explanation:
h(x) = 6 - X
Let x = -10
h(-10) = 6- -10
= 6+ 10
= 16
The triangles are ACD and XYZ
Congruent segments means that they have the same length.
Then CD = XY, AC = ZX and AD = ZY
This is the option B. from the list.
Answer :<span>B.CD=XY,DA=YZ,AC=ZX</span>
