The ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
<u>Explanation:</u>
We need to find the time at which the ball will be at height 19 feet.
Equation:
h = 3 + 34t - 16t²
19 = 3 + 34t - 16t²
16 = -16t² + 34t
-16t² + 34t - 16 = 0
On solving the equation, we get
t1 = 0.7 s and t2 = 1.42s
Therefore, the ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
Evaluate abs(2 x - 7) - 3 where x = 2:abs(2 x - 7) - 3 = abs(2 2 - 7) - 3
2×2 = 4:abs(4 - 7) - 3
4 - 7 = -3:abs(-3) - 3
Since -3<=0, then abs(-3) = 3:3 - 3
3 - 3 = 0:Answer: 0
7(6) x 7(5) = 7(6+5) = 7(11) is the simplified answer.
Thank you!
P(x) = √x
for x = 0 → √0 = 0 and p(0) = 0
for x = 1.44 → √1.44 =1.2 and p(1.44) = 1.2
for x = 2.25 → √2.25 = 1.5 and p(2.25) = 1.5
for x = 3.24 → √3.24 = 1.8 and p(3.24) = 1.8
for x = 4.41 → √4.41= 2.1 and p(4.41) = 2.1
for x = 5.29 → √5.29 = 2.3 and p(5.29) = 2.3
Area=1/2 times base times height
for equilateral triangle, if base is x, then height is (x√3)/2
so if base is 6 then height is (6√3)/2=3√3
area is 1/2 times 6 times3√3=3 times 3√3=9√3 sqare inches
