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3 years ago

Write the equation of the parabola y=x^2 + 8x - 51

1 answer:

7 0

You already have the equation! If you want it in vertex form, do this:

y = x^2 + 8x + 16 - 16 - 51, or y = x^2 + 8x + 16 -67, or
y = (x+4)^2 - 67.
The vertex is at (-4,67).

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

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago

Someone help me please I don't know how to do this ! :(

Ivanshal [37]

$4^{x+1}=21\\
x+1=\log_421\\
x=\log_421-1\\
x=\log_421-\log_44\\
x=\log_4\dfrac{21}{4}\\
x=\dfrac{\ln \dfrac{21}{4}}{\ln 4}\approx 1.1962$

5 0

3 years ago

The perimeter of a square is 20 feet. How long is each side?

Firdavs [7]

Answer:

5ft

Step-by-step explanation:

all sides of a square are equal and there are 4 sides so divide 20 by 4 and you get 5 ft

5 0

2 years ago

Read 2 more answers

Ann is adding water to a swimming pool at a constant rate. The table below shows the amount of water in the pool after different

N76 [4]

Answer:

(a) As time increases, the amount of water in the pool increases.

11 gallons per minute

(b) 65 gallons

Step-by-step explanation:

From inspection of the table, we can see that as time increases, the amount of water in the pool increases.

We are told that Ann adds water at a constant rate. Therefore, this can be modeled as a linear function.

The rate at which the water is increasing is the rate of change (which is also the slope of a linear function).

Choose 2 ordered pairs from the table:

$\textsf{let}\:(x_1,y_1)=(8, 153)$

$\textsf{let}\:(x_2,y_2)=(12,197)$

Input these into the slope formula:

$\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{197-153}{12-8}=\dfrac{44}{4}=11$

Therefore, the rate at which the water in the pool is increasing is:

11 gallons per minute

To find the amount of water that was already in the pool when Ann started adding water, we need to create a linear equation using the found slope and one of the ordered pairs with the point-slope formula:

$y-y_1=m(x-x_1)$

$\implies y-153=11(x-8)$

$\implies y-153=11x-88$

$\implies y=11x-88+153$

$\implies y=11x+65$

When Ann had added no water, x = 0. Therefore,

$y=11(0)+65$

$y=65$

So there was 65 gallons of water in the pool before Ann starting adding water.

3 0

2 years ago

Read 2 more answers

Which graph correctly solves the system of equations below? y=-x^2+1 y=x^2-4

KIM [24]

Red line shows
y=x^2-4 , cause it's going through point (0,-4)

So blue line shows
y= -x^2+1 Point (1,0) belong to it , moreover x factor is negative , so the arms of parabola will go down

Have nice evenig :)
Greetings From Poland.

5 0

3 years ago