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3 years ago

Which equation is y= 2x ^2 - 8x +9 rewritten in vertex form.

Mathematics

2 answers:

REY [17]3 years ago

5 0

vertex form wow tht's advance um i'll say what you wrote up there is the answer

Licemer1 [7]3 years ago

4 0

y= 2x ^2 - 8x +9

y = a(x - h)2 + k, where (h, k) is the vertex of the parabola
so
y= 2x ^2 - 8x + 9
y= 2x ^2 - 8x + 8 + 1
y = 2(x^2 - 4x - 4) + 1
y = 2(x - 2)^2 + 1 ....<---------vertex form

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A Ferris Wheel is built such that the height in feet above ground of a seat on the wheel at time t seconds can be modeled by h(t

MAVERICK [17]

Answer:

at time, t = 8 seconds and t = 24 seconds Ferris Wheel be 53 feet above the ground

Step-by-step explanation:

Data provided in the question:

height in feet above ground of a seat on the wheel at time t seconds is

modeled as

h(t) = $53 + 50\sin((\frac{\pi t}{16} - \frac{\pi}{2})$

now,

at height 53 above the ground, we get the equation as:

53 = $53 + 50\sin(\frac{\pi t}{16} - \frac{\pi}{2})$

$50\sin(\frac{\pi t}{16} - \frac{\pi}{2})$ = 53 - 53

$\sin(\frac{\pi t}{16} - \frac{\pi}{2})$ = 0

also,

sin(0) = 0

and,

sin(π) = 0

therefore,

$(\frac{\pi t}{16} - \frac{\pi}{2})$ = 0

$\frac{\pi t}{16} = \frac{\pi}{2}$

t = 8 seconds

and,

$(\frac{\pi t}{16} - \frac{\pi}{2})$ = π

$\frac{\pi t}{16}= \pi + \frac{\pi}{2}$

$\frac{\pi t}{16}= \frac{3\pi}{2}$

t = 24 seconds

Hence,

the at time, t = 8 seconds and t = 24 seconds Ferris Wheel be 53 feet above the ground

7 0

3 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$