Answer: 10 squares.
Step-by-step explanation:
We can do it with integrals.
If we take the bottom vertex as the point (0,0), the top right vertex as the point (0,4) and the left vertex as the point (5, 6)
then the area of the triangle is the area enclosed by two lines between the values x = 0 and x = 5.
the two lines are:
the bottom is the one that passes through (0,0) and (5,6)
f(x) = y = s*x (because it passes through the point (0,0))
and the slope is s = 6/5.
so f(x) = y = (6/5)*x.
The top line is the one that passes through (0,4) and (5,6)
the y-intercept is b = 4, and the slope is:
s = (6 - 4)/(5 - 0) = 2/5.
g(x) = y = (2/5)*x + 4.
Now, the area enclosed for the triangle is equal to:
this is equal to:
= (1/2)(-4/5)*(5)^2 + 4*5 - 0 = 10
The area is 10 squares
Answer:
f(x) is concave up whenever:
B. 3x²−10 is positive
f(x) is concave down whenever:
A. 3x²−10 is negative
The points of inflection of f(x) are the same as:
B. the zeros of 3x²−10
Step-by-step explanation:
Given the function f(x) = 1 / (x²+10)
We can determine the concavity by finding the second derivative.
If
f"(x) > 0 ⇒ f(x) is concave up
If
f"(x) < 0 ⇒ f(x) is concave down
Then
f'(x) = (1 / (x²+10))' = -2x / (x²+10)²
⇒ f"(x) = -2*(10-3x²) / (x²+10)³
if f"(x) = 0 ⇒ -2*(10-3x²) = 0 ⇒ 3x²-10 = 0
f(x) is concave up whenever 3x²−10 > 0
f(x) is concave down whenever 3x²−10 < 0
The points of inflection of f(x) are the same as the zeros of 3x²-10
it means that 3x²-10 = 0
Answer:
Step-by-step explanation:
pq/3-q/2
plug in numbers
3x4/3-4/2
12/3-2
4-2
answer is 2
Is there answer choices ?
Answer:
where is the rest of the question?
Step-by-step explanation:
