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2 years ago

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Peter worked 2 part-time jobs and earned $189.00 for working a total of 24 hours at both jobs. He is paid $6.00 per hour at Job

A abs $9.00 per hour at Job B. How many hours did Peter work at Job A?

1 answer:

timama [110]2 years ago

8 0

You use the substitution method simultaneously. He worked 9h at job A

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What is the value of the ratio 12 1/2 14 3/4 feet

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3 years ago

Solve using algebraic equation: <br> 5sin2x=3cosx<br><br> (No exponents)

DaniilM [7]

$5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x$

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

$10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0$

Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$ , so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$ . More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$ , the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$ .

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3 years ago

1. The position of a particle moving along a coordinate axis is given by: s(t) = t^2 - 5t + 1. a) Find the speed of the particle

zimovet [89]

Answer: $\left | 2t-5\right |,\ 2,\ 2t-5$

Step-by-step explanation:

Given

Position of the particle moving along the coordinate axis is given by

$s(t)=t^2-5t+1$

Speed of the particle is given by

$\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=\left | 2t-5\right |$

Acceleration of the particle is

$\Rightarrow a=\dfrac{dv}{dt}\\\\\Rightarrow a=2$

velocity can be negative, but speed cannot

$\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=2t-5$

3 0

3 years ago