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vodka [1.7K]
1 year ago
6

Explain how complex fractions can be used to solve problems involving ratios

Mathematics
1 answer:
Kisachek [45]1 year ago
8 0
When you have ratios and some unknowns you can create complex fractions from them.Bring them to the same denominator and solve for X.
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Find the sine function that is represented in the graph.
meriva

Answer:

Choice D). f(x)=20\sin(4x) is correct.

Step-by-step explanation:

Given function graph is sinusoidal so let's compare with formula f(x)=A\sin(Bx-C)+D

We know that amplitude is the height from the center line to the peak (or to the trough). From graph we can see that height from the center line to the peak is 20

So amplitude A=20

In that formula, period is given by \frac{2\pi}{B}

From graph we see that period is \frac{\pi}{2}

So both must be equal

\frac{2\pi}{B}=\frac{\pi}{2}

\frac{2}{B}=\frac{1}{2}

cross multiplying them gives

B=4

Clearly there is no shift so C and D are 0

Now plug these values into formula f(x)=A\sin(Bx-C)+D

f(x)=20\sin(4x-0)+0

f(x)=20\sin(4x)

Hence choice D is correct.

6 0
1 year ago
If line b is perpendicular to line a, and line c is perpendicular to line a, what is the slope of line c?
slega [8]

we know that

If line b is perpendicular to line a, and line c is perpendicular to line a,

then

line b and line c are parallel

and two lines parallel have the same slope

so

<u>Find the slope of the line b</u>

Let

A(-3,-2)\\B(2,3)

The formula to calculate the slope between two points is equal to

m=\frac{(y2-y1)}{(x2-x1)}

(x1,y1)=(-3,-2)\\(x2,y2)=(2,3)

substitute

m=\frac{(3+2)}{(2+3)}

m=\frac{(5)}{(5)}

m=1

therefore

<u>the answer is</u>

<u>the slope of the line c is</u>

mc=1

8 0
2 years ago
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Angela shared a cab with her friends. When they arrived at their destination, they evenly divided the \$j$jdollar sign, j fare a
irakobra [83]
Y=3x+5 is the answer to this problem hope it helped
3 0
2 years ago
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Isa and Jack have part-time jobs.
enot [183]
By process of elimination and plugging in the math we can say that Jack earns 8.25 an hour by multiplying 2*8.25=16.50. Isa must then earn 9.50, therefore she earns a greater hourly pay then Jack as a result 
3 0
2 years ago
Use Definition 7.1.1, DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = ∞ e−
poizon [28]

f(t)=\begin{cases}\cos t&\text{for }0\le t

The Laplace transform is then

\mathcal L_s\{f(t)\}=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=\int_0^\pi e^{-st}\cos t\,\mathrm dt

Let I denote the integral we want to compute. Integrating by parts, setting

u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt

\mathrm dv=\cos t\,\mathrm dt\implies v=\sin t

gives

\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\int_0^\pi e^{-st}\sin t\,\mathrm dt

Integrate by parts again, setting

u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt

\mathrm dv=\sin t\,\mathrm dt\implies v=-\cos t

Then

\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\left(-e^{-st}\cos t\bigg|_{t=0}^{t=\pi}-s\int_0^\pi e^{-st}\cos t\,\mathrm dt\right)

I=e^{-st}(\sin t-s\cos t)\bigg|_{t=0}^{t=\pi}-s^2I

(s^2+1)I=s(e^{-\pi s}+1)

I=\dfrac s{s^2+1}(e^{-\pi s}+1)

7 0
2 years ago
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