Answer:
the last one
Step-by-step explanation:
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:
26
hope this heps and pls mark brainliest :))
Step-by-step explanation:
Answer:
y/x = 2.5
Step-by-step explanation:
y ∝ x
change ∝ to = by adding constant k or c
y = kx
Divide both sides by x
y/x = k, also k = y/x
Input the values
k = (7 1/2)/3. Change 7 1/2 to improper fraction to give 15/2
k = (15/2)/3
k = 
÷ 3
Change the division sign to multiplication which will change 3 to its inverse(1/3)
k = × 
k = 
Divide the numerator and denominator by 3 to get 5/2
k = 5/2
From y/x = k,input k into the eqn
y/x = 5/2
y/x = 2.5
Answer:
y = -2x^2(x - 3)
Step-by-step explanation:
<em><u>Preliminary Remark</u></em>
If a cubic is tangent to the x axis at 0,0
Then the equation must be related to y = a*x^2(x - h)
<em><u>(3,0)</u></em>
If the cubic goes through the point (3,0), then the equation will become
0 = a*3^2(3 - h)
0 = 9a (3 - h)
0 = 27a - 9ah
from which h = 3
<em><u>From the second point, we get</u></em>
4 = ax^2(x - 3)
4 = a(1)^2(1 - 3)
4 = a(-2)
a = 4 / - 2
a = -2
<em><u>Answer</u></em>
y = -2x^2(x - 3)
