The formula for average velocity between two times t1 and t2 of the position function f(x) is (f(t2)-f(t1)) / (t2-t1)
Plugging the values in for the first time period we get (f(2.5)-f(2)) / (2.5-2)
=> (f(2.5)-f(2)) / 0.5
f(2) will be the same for all 4 time periods and is
48(2)-16(2)^2 = 32
Now we plugin the other values
f(2.5) = 48(2.5)-16(2.5)^2 = 20
f(2.1) = 48(2.1)-16(2.1)^2 = 30.25
etc.
f(2.05) = 31.16
f(2.01) = 31.8384
Now plug these values into the formula
(20-32)/0.5 = -24
(30.25-32)/0.1 = -17.5
etc.
= -16.8
= -16.16
Final answer:
2.5s => -24 ft/s
2.1s => -17.5 ft/s
2.05 => -16.8 ft/s
2.01 => -16.16 ft/s
Hope I helped :)
Answer:
Yes
Step-by-step explanation:
All rhombuses are parallelograms, but not all parallelograms are rhombuses. All squares are rhombuses, but not all rhombuses are squares. The opposite interior angles of rhombuses are congruent. Diagonals of a rhombus always bisect each other at right angles.
Answer:
The sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is <u>169</u>
Step-by-step explanation:
Given : the difference of the squares of the numbers is 5 and the product of the numbers is 6.
We have to find the sum of the squares of two numbers whose difference and product is given using given identity,
Since, given the difference of the squares of the numbers is 5 that is 
And the product of the numbers is 6 that is 
Using identity, we have,
Substitute, we have,
Simplify, we have,
Thus, the sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is 169
Answer:
Step-by-step explanation:
Given the formula for calculating the distance travelled expressed as;
s=1/2at^2
Given
a = 3
t = 10
Required
lower bound of s
Substitute the given values into the equation;
s=1/2at^2
S = 1/2(3)(10)^2
S = 1/2 * 3 * 100
S = 3 * 50
S = 150
Hence the lower bound of distance S is 150
