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3 years ago

What are the answers cause I keep getting different answers when I check my work

Mathematics

1 answer:

zmey [24]3 years ago

8 0

Answer:First one is parallel

Step-by-step explanation:

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1/2x = 17.31 I need help finding out what can equals??

kondor19780726 [428]

Solve for xx by simplifying both sides of the equation, then isolating the variable.x=34.62

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3 years ago

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100 points!!! Pre calculus. I need helpppppppppp

Fantom [35]

Answer:

We have function,

$y = 3 - 6 \sin {}^{} (2x + \frac{\pi}{2} )$

Standard Form of Sinusoid is

$y = - 6 \sin(2x + \frac{\pi}{2} ) + 3$

Which corresponds to

$y = a \sin(b(x + c)) + d$

where a is the amplitude

2pi/b is the period

c is phase shift

d is vertical shift or midline.

In the equation equation, we must factor out 2 so we get

$y = - 6(2(x + \frac{\pi}{4} )) + 3$

Also remeber a and b is always positive

So now let answer the questions.

a. The period is

$\frac{2\pi}{ |b| }$

$\frac{2\pi}{ |2| } = \pi$

So the period is pi radians.

b. Amplitude is

$| - 6| = 6$

Amplitude is 6.

c. Domain of a sinusoid is all reals. Here that stays the same. Range of a sinusoid is [-a+c, a-c]. Put the least number first, and the greatest next.

So using that rule, our range is [6+3, -6+3]= [9,-3] So our range is [-3,9].

D. Plug in 0 for x.

$3 - 6 \sin((2(0) + \frac{\pi}{2} )$

$3 - 6 \sin( \frac{\pi}{2} )$

$3 - 6(1)$

$3 - 6$

$= - 3$

So the y intercept is (0,-3)

E. To find phase shift, set x-c=0 to solve for phase shift.

$x + \frac{\pi}{4} = 0$

$x = - \frac{\pi}{4}$

Negative means to the left, so the phase shift is pi/4 units to the left.

f. Period is PI, so use interval [0,2pi].

Look at the graph above,

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3 years ago

What is 238 rouned to the nearst houndred

Alexeev081 [22]

200 because:
250 or higher is 300
249 or lower is 200

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3 years ago

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Chuck gets a 7% raise this year.if chucks salary is equal to "t", write the equation to determine Chucks's new salary

saw5 [17]

Hello

salaire = 1.07 t

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3 years ago

A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the

amm1812

Answer:

$s(t)=(ln|7|+ln|2|)\,ft$

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

$a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\$

Integrating both sides w.r.to t

$v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\$

Using initial value

$v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}$

We have to calculate the distance covered in time interval [0,5], so:

$\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft$

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3 years ago