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3 years ago

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A ball is thrown from 4 feet off the ground with a vertical velocity of 30 feet per second. Write an equation to model the heigh

t of the ball: h = -16t 2 + a0 t + a1 Complete the table.

1 answer:

olga55 [171]3 years ago

5 0

A0 = 30t
a1 = 4

This is the standard model for h and t when discussing physics.

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Find the percent of the area under the density curve where x is more than 3. <br>

Gekata [30.6K]

Answer:

The percent of the area under the density curve where $x$ is more that 3 is 25 %.

Step-by-step explanation:

Since the density curve is a linear function, the area under the curve can be calculated by the geometric formula for a triangle, defined by the following expression:

$A = \frac{1}{2}\cdot (x_{f} - x_{o})\cdot (y_{f}-y_{o})$ (1)

Where:

$A$ - Area, in square units.

$x_{f} - x_{o}$ - Base of the triangle, in units.

$y_{f} - y_{o}$ - Height of the triangle, in units.

The percent of the area is the ratio of triangle areas under the density curve multiplied by 100 per cent, that is:

$x = \frac{\frac{1}{2}\cdot (5-3)\cdot (0.25) }{\frac{1}{2}\cdot (5-1)\cdot (0.5) }\times 100\,\%$

$x = 25\,\%$

The percent of the area under the density curve where $x$ is more that 3 is 25 %.

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3 years ago

(-w^3 + 8w^2 – 3w) + (-8w^2 + w + 3) =

kherson [118]

The expression simplified is -w^3-2w+3

I hope this helped

4 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

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3 years ago

Solve for x. Show your work.<br> -1/2x < -12x

cricket20 [7]

Answer: x > 24

Step-by-step explanation: -12x/-1/2x = 24. 24 is the given when dividing. The sign is flipped after dividing negatives.

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3 years ago

Read 2 more answers

In the parabola y=(x+1)^2+2

Svet_ta [14]

Direction: Opens Up

Vertex: (1,2)

Focus: (1, 9/4)

Axis of Symmetry: x = 1

Directrix: y = 7/4

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3 years ago