1.) set up a w_k_u chart
2.) in the w column write in on top and mi in the bottom like this
w_K_U
in_
mi_
then in the k column write 2 for in and 15 for mi
W_K_U
in_2_
mi_15_
then in the U column, you write 7 for in and x for mi
W_K_U
in_2_7
mi_15_x
then cross multiply: 2x = 15(7)
multiply 15 times 7 which is 105
2x=105
divide 2x by 2
x=105
answer 105
Answer:
2. a and b only.
Step-by-step explanation:
We can check all of the given conditions to see which is true and which false.
a. f(c)=0 for some c in (-2,2).
According to the intermediate value theorem this must be true, since the extreme values of the function are f(-2)=1 and f(2)=-1, so according to the theorem, there must be one x-value for which f(x)=0 (middle value between the extreme values) if the function is continuous.
b. the graph of f(-x)+x crosses the x-axis on (-2,2)
Let's test this condition, we will substitute x for the given values on the interval so we get:
f(-(-2))+(-2)
f(2)-2
-1-1=-3 lower limit
f(-2)+2
1+2=3 higher limit
according to these results, the graph must cross the x-axis at some point so the graph can move from f(x)=-3 to f(x)=3, so this must be true.
c. f(c)<1 for all c in (-2,2)
even though this might be true for some x-values of of the interval, there are some other points where this might not be the case. You can find one of those situations when finding f(-2)=1, which is a positive value of f(c), so this must be false.
The final answer is then 2. a and b only.
Answer: C
Step-by-step explanation:
<h3>
Answer: Choice D) </h3>
Work Shown:

We must require that
and
to avoid having 0 in the denominator. This is why choice D is the answer.
Answer:

Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable.