<h2>
Answer:</h2>
<u>A direct variation equation is of the form y = m⋅x for some constant value m</u>.
<u>For a direct variation equation passing through</u>
(x,y) = (-11,13)
13 = m × - 11
→ m = - 13/11.
so, as y = m⋅x
y = - 13/11x
<u>Hence, the direct variation equation is [C] - 13/11x</u>.
Answer:
Step-by-step explanation:
As you can see from the graph I attached you, the possible solutions in the interval from 0 to 2π are approximately:
So, it's useful to solve the equation too, in order to verify the result:
Taking the inverse sine of both sides:
Using this result we can conclude the solutions in the interval from 0 to 2π are approximately:
Term to be added is: + 121/4
t^2 + 11t + 121/4
(t + 11/2) (t + 11/2)
Answer:
Step-by-step explanation:
Answer: Options F, E, C.
Given : function f(x)=2(x-4)^5.
To find : What are the characteristics of the function .
Solution : We have given that f(x)=2(x - 4)^5.
By the End Point behavior : if the degree is even and leading coefficient is odd of polynomial of function then left end of graph goes down and right goes up.
Since , Option F is correct.
It has degree 5 therefore, function has 5 zeros and atmost 4 maximua or minimum.
Option E is also correct.
By transformation rule it is vertical stretch and shift to right (B )
Therefore, Option F , E , C are characteristics of the function .
Answer:
C
Step-by-step explanation:
If you look at your y value which would be your Output you can see the number 4 repeated multiple times
Therefore leading up to the answer that there is more than what output.
The input is the number you feed into the expression, and the output is what you get after the look-up work or calculations are finished.
Hope this helps :)
