13. A catapult launches a boulder with an upward velocity of 148 ft/s. The height of the boulder (h) in feet after t seconds is given by the function h = –16t² + 148t + 30. How long does it take the boulder to reach its maximum height? What is the boulder’s maximum height? Round to the nearest hundredth, if necessary.
A) 9.25 s; 30 ft
B) 4.63 s; 640.5 ft
C) 4.63 s; 1,056.75 ft
D) 4.63 s; 372.25 ft

16. Solve the equation using the Zero Product Property.
(2x – 4)(2x – 1) = 0
A) 2, –1over2
B) 2, 1over2
C) –2, 2
D) –2, 1over2

32t + 148 = 0148 = 32t4.625 = tt ≈ 4.63h = -16 • (4.625)^2 + 148 • 4.625 + 30 = 372.25
ANSWER: is D. 4.63 sec; 372.25 ft
2.) 2x - 4 = 0 and 2x - 1 = 0x = {4/2, ½}x = {2,½} ANSWER: is B 2, 1 over 2

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lidaralbany lidaralbany · Answer 2 · 2019-04-04T07:46:51+02:00

13. Answer: The height will be maximum 372.25 ft at 4.63 s.

Explanation:

Given that,

Upward velocity v = 148 ft/s

Function

[tex]h = -16t^2+148t+30[/tex].....(I)

On differentiate

[tex]\dfrac{dh}{dt}=-32t+148[/tex]....(II)

for maximum height,

[tex]\dfrac{dh}{dt}=0[/tex]

Put the value of [tex]\dfrac{dh}{dt}[/tex] in equation (II)

[tex]-32t+148=0[/tex]

[tex]t = \dfrac{148}{32}[/tex]

[tex]t = 4.63\ s[/tex]

The maximum height is

Put the value of t in equation(I)

[tex]h = -16\times(4.63)^2+148\times4.63+30[/tex]

[tex]h = 372.25 ft[/tex]

Hence, The height will be maximum 372.25 ft at 4.63 s.

16. Answer: The value of x is 2, [tex]\dfrac{1}{2}[/tex].

Explanation:

Given that,

[tex](2x-4)(2x-1)=0[/tex]

Using zero product property

[tex]2 x-4=0[/tex]

[tex]x = 2[/tex]

And,

[tex]2x-1=0[/tex]

[tex]x = \dfrac{1}{2}[/tex]

Hence, The value of x is 2, [tex]\dfrac{1}{2}[/tex].