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 :)
Correct Question: If m∠JKM = 43, m∠MKL = (8x - 20), and m∠JKL = (10x - 11), find each measure.
1. x = ?
2. m∠MKL = ?
3. m∠JKL = ?
Answer/Step-by-step explanation:
Given:
m<JKM = 43,
m<MKL = (8x - 20),
m<JKL = (10x - 11).
Required:
1. Value of x
2. m<MKL
3. m<JKL
Solution:
1. Value of x:
m<JKL = m<MKL + m<JKM (angle addition postulate)
Therefore:
Solve for x
Subtract 8x from both sides
Add 11 to both sides
Divide both sides by 2
2. m<MKL = 8x - 20
Plug in the value of x
m<MKL = 8(17) - 20 = 136 - 20 = 116°
3. m<JKL = 10x - 11
m<JKL = 10(17) - 11 = 170 - 11 = 159°
Answer:
See explanation
Step-by-step explanation:
3(x + 4) + 2 = 2 + 5(x – 4)
Step 1: distributive property
3(x + 4) + 2 = 2 + 5(x – 4)
3x + 12 + 2 = 2 + 5x - 20
Step 2: collect like terms
3x + 12 + 2 = 2 + 5x - 20
3x + 14 = 5x - 18
Step 3: Addition property of equality
3x + 14 = 5x - 18
3x + 14 + 18 = 5x - 18 + 18
3x + 32 = 5x
Step 4: subtraction property of equality
3x + 32 - 3x = 5x - 3x
32 = 2x
Step 5: division property of equality
32 = 2x
32/2 = 2x/2
16 = x
x = 16
Answer:
(2. 16/24) (3. 0 (because there are no green shirts))
Step-by-step explanation:
Hope it helps!
