Answer:
I think the answer is B
Step-by-step explanation:
For this case what we should do is evaluate the function in each of the points from 2 to 7 and then find the average.
We have then:
f (x) = ln (3x) +2.5.
f (2) = ln (3 * 2) + 2.5 = 4.291759469
f (3) = ln (3 * 3) + 2.5 = 4.697224577
f (4) = ln (3 * 4) + 2.5 = 4.98490665
f (5) = ln (3 * 5) + 2.5 = 5.208050201
f (6) = ln (3 * 6) + 2.5 = 5.390371758
f (7) = ln (3 * 7) + 2.5 = 5.544522438
Then the average is:
Average = (4.291759469 + 4.697224577 + 4.98490665 + 5.208050201 + 5.390371758 + 5.544522438) / (6) = 5.019472516
nearest hundredth
5.02
Answer:
the average rate of change of f (x) from 2 to 7 is 5.02
You can find the area of Bonnue's backyard by comparing the hypotenuse of the garden to the hypotenuse of the back yard. If the hypotenuse of the garden is 10 (with the side lengths being 6, 8 and 10 - the longest is always the hypotenuse) and the hypotenuse of the back yard is 30, this is a scale factor of 3 (3 times longer).
This means the other two sides would also be 3 times longer.
6 yards x 3 = 18
8 yards x 3 = 24
To find the area using these dimensions, you will use the formula for finding the area of a triangle.
A = 1/2bh
A = 1/2 x 18 x 24
A = 216 square yards
The area of the backyard is 216 square yards.
bakers a and d
[6:4 is the same as 3:2]
(hope this helps, if so, please give brainliest x)
<u>Answer:</u>
a) 3.675 m
b) 3.67m
<u>Explanation:</u>
We are given acceleration due to gravity on earth =
And on planet given = 
A) <u>Since the maximum</u><u> jump height</u><u> is given by the formula </u>
Where H = max jump height,
v0 = velocity of jump,
Ø = angle of jump and
g = acceleration due to gravity
Considering velocity and angle in both cases
Where H1 = jump height on given planet,
H2 = jump height on earth = 0.75m (given)
g1 = 2.0
and
g2 = 9.8
Substituting these values we get H1 = 3.675m which is the required answer
B)<u> Formula to </u><u>find height</u><u> of ball thrown is given by </u>
which is due to projectile motion of ball
Now h = max height,
v0 = initial velocity = 0,
t = time of motion,
a = acceleration = g = acceleration due to gravity
Considering t = same on both places we can write
where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations
substituting h2 = 18m, g1 = 2.0 and g2 = 9.8
We get h1 = 3.67m which is the required height
