Replace the second equation y=3x+4 into the first one:
2(3x+4)+5x=30
Now solve for x:
6x+8+5x=30
11x=30-8
11x=22
x=22/11
x=2
Finally, replace x=2 into y=3x+4
y=3*2+4
y=6+4 = 10
So the anwer is x=2, y=10
<span>The number of x-intercepts that appear on the graph of the function
</span>f(x)=(x-6)^2(x+2)^2 is two (2): x=6 (multiplicity 2) and x=-2 (multiplicity 2)
Solution
x-intercepts:
f(x)=0→(x-6)^2 (x+2)^2 =0
Using that: If a . b =0→a=0 or b=0; with a=(x-6)^2 and b=(x+2)^2
(x-6)^2=0
Solving for x. Square root both sides of the equation:
sqrt[ (x-6)^2] = sqrt(0)→x-6=0
Adding 6 both sides of the equation:
x-6+6=0+6→x=6 Multiplicity 2
(x+2)^2=0
Solving for x. Square root both sides of the equation:
sqrt[ (x+2)^2] = sqrt(0)→x+2=0
Subtracting 2 both sides of the equation:
x+2-2=0-2→x=-2 Multiplicity 2
Answer:
RU is 8
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
For any of the functions described above, the only way any of those could be functions is that there has to be a difference value for each x, unless it is the same x-value. If the x is mentioned twice, that is fine, as long as the y point is also the same. If it is different, it is not a function.
Answer:
min: 8
max:28
med: 22
1st quartile: 12
3rd quartile: 26
hope this helps :)))
Step-by-step explanation:
