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VMariaS [17]
3 years ago
9

We're working with single variable equations and I'm okay at them, but I'm rusty when it comes to using it with fractions. If an

yone could walk me through this problem I'd really appreciate it.
\frac{v}{3}  + 2 =  \frac{4}{5}
​
Mathematics
1 answer:
jasenka [17]3 years ago
3 0

Answer: v=-18/5

Step-by-step explanation:

Assuming you want to solve for v, we need to use our algebraic properties.

\frac{v}{3}+2=\frac{4}{5}                 [subtract both sides by 2]

\frac{v}{3}=-\frac{6}{5}                    [multiply both sides by 3]

v=-\frac{18}{5}

Now, we know that v=-18/5.

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(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr
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f(t)=\begin{cases}0&\text{for }t

Recall that

u(t)=\begin{cases}0&\text{for }t

Take it one piece at a time. For t\ge0, we can scale u(t) by -5:

-5u(t)=\begin{cases}0&\text{for }t

If we shift the argument by 1 and scale by -5, we have

-5u(t-1)=\begin{cases}0&\text{for }t

so if we subtract this from -5u(t), we'll end up with

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For the next piece, we can add another scaled and shifted step like

-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

so that

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For the last piece, we add one more term:

6u(t-7)=\begin{cases}0&\text{for }t

and so putting everything together, we get f(t):

f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)
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