Step-by-step explanation:
I assume that "ground" is at 0 ft height. which is in an actual scenario not airways the case.
y = -16x² + 64x + 89
shows us that the tower is 89 ft tall (the result for x = 0, at the start).
anyway, if the original assumption is correct, then we need to solve
0 = -16x² + 64x + 89
the general solution for such a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/)2a)
in our case
a = -16
b = 64
c = 89
x = (-64 ± sqrt(64² - 4×-16×89))/(2×-16) =
= (-64 ± sqrt(4096 + 5696))/-32 =
= (-64 ± sqrt(9792))/-32
x1 = (-64 + 98.95453501...)/-32 = -1.092329219... s
x2 = (-64 - 98.95453501...)/-32 = 5.092329219... s
the negative solution for time is but useful here (it would be the time calculated back to ground at the start).
so, x2 is our solution.
the rocket hits the ground after about 5.09 seconds.
Answer:
The best estimate for 43x78 is 40x80
Step-by-step explanation:
43 rounded to the nearest ten is 40.
78 rounded to the nearest ten is 80.
The answer to 40x80 is 3200
Answer:
http://www.classzone.com/science_book/mls_grade6_FL/540_547.pdf
Step-by-step explanation:
The correct answer is: [B]: " 25 a²⁵ b²⁵ " .
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<span>Explanation:
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Given the expression:
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</span>→ " (−5a⁵b⁵)² (a³b³)⁵ " ; Simplify.
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Let us being by examining:
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→ "(−5a⁵b⁵)² " .
→ "(−5a⁵b⁵)² = (-5)² * (a⁵)² * (b⁵)² = (-5)(-5) * a⁽⁵ˣ²⁾ * b⁽⁵ˣ²⁾ = 25a⁽¹⁰⁾b⁽¹⁰⁾ ;
{Note the following properties of exponents:
(xy)ⁿ = xⁿ * yⁿ ;
(xᵃ)ᵇ = x⁽ᵃ * ᵇ) ;
(xᵃ) * (xᵇ) = x⁽ᵃ ⁺ ᵇ⁾ .}.
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Then, we examine:
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→ "(a³b³)⁵ " .
→ "(a³b³)⁵ = a⁽³ˣ⁵⁾b⁽³ˣ⁵⁾ = a⁽¹⁵⁾b⁽¹⁵⁾ .
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So: " (−5a⁵b⁵)² (a³b³)⁵ = (-5)a⁽¹⁰⁾b⁽¹⁰⁾ * a⁽¹⁵⁾b⁽¹⁵⁾ " ;
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Now, we simplify:
→ " 25a⁽¹⁰⁾b⁽¹⁰⁾ * a⁽¹⁵⁾b⁽¹⁵⁾ " ;
→ " 25a⁽¹⁰⁾b⁽¹⁰⁾ * a⁽¹⁵⁾b⁽¹⁵⁾ ;
= 25a⁽¹⁰⁾ a⁽¹⁵⁾b⁽¹⁰⁾ b⁽¹⁵⁾ ;
= 25a⁽¹⁰ ⁺¹⁵⁾ b⁽¹⁰⁺¹⁵⁾ ;
= 25a⁽²⁵⁾ b⁽²⁵⁾ ;
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→ which is: Answer choice: [B]: " 25 a²⁵ b²⁵ " .
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Answer:18
Step-by-step explanation: 18=2.3.3
