<span>Formula: H(t) = 56t – 16t^2
</span>
H(t) = - 16t^2 + 56t
<span><span>A.
</span>What is the height of the ball after 1 second? H
(1) = 56(1) – 16(1) ^2 = 40 pt.</span>
<span><span>B.
</span>What is the maximum height? X = - (56)/2(- 16) =
1.75 sec h (1.75) = 56(1.75) – 16(1.75) ^2 h (1.75) = 49ft.</span>
<span><span>C.
</span><span>After how many seconds will it return to the
ground? – 16t^2 + 56t = 0 - 8t =0 t = 0</span></span>
<span><span>-
</span><span>8t (2 + - 7) = 0 2t – 7 = 0 t = 7/2
Ans: 3.5 seconds</span></span>
Answer:
Step-by-step explanation:
<h3>AP given</h3>
<h3>To find</h3>
<h3>Solution</h3>
Common difference
<u>Difference of first two</u>
- d = (a -b) - (a + b) = -2b
<u>Difference of second two</u>
<u>Difference of last two</u>
<u>Now comparing d:</u>
- -2b = ab - (a - b)
- ab - a = - 3b
- a(1 - b) = 3b
- a = 3b/(1 - b)
and
- a/b - ab = -2b
- a(1/b - b) = -2b
- a = 2b²/(b² - 1)
<u>Eliminating a:</u>
- 2b²/(b² - 1) = 3b/(1 - b)
- 2b/(b+1) = -3
- 2b = -3b - 3
- 5b = - 3
- b = -3/5
<u>Finding a:</u>
- a = 3b/(1 - b) =
- 3*(-3/5) *1/(1 - (-3/5)) =
- -9/5*5/8 =
- -9/8
<u>So the first term is:</u>
- a + b = -3/5 - 9/8 = -24/40 - 45/40 = - 69/40
<u>Common difference:</u>
<u>The 6th term:</u>
- a₆ = a₁ + 5d =
- -69/40 + 5*6/5 =
- -69/40 + 240/40 =
- 171/40 = 4 11/40
A quadrilateral has 4 vertices and the sum of the measures of the interior angles is 360 degrees. There cannot be 4 diagonals in a quadrilateral.
=12-5=7
=21-12=9
when u look at they have a diffrence of 2 so the answer will have a diffrence of 2 =11
Answer:
false bc the y intercept is in a different place which in return would alter the entire equation
