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larisa86 [58]
2 years ago
13

Y=4x+1 complete the table

Mathematics
1 answer:
Tom [10]2 years ago
3 0

<em>Please</em><em> </em><em>refer</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>image</em><em> </em><em>for</em><em> </em><em>the</em><em> </em><em>answer</em>

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A number is chosen at random from 1 to 10. Find the probability of selecting factors of 4 and factors
Lena [83]
The ANWSER is 2/4 I hope I helped
8 0
3 years ago
Write the equation of the line with the given slope and the given point.<br><br> y = ?
Naddika [18.5K]

Answer:

y=x+8

Step-by-step explanation:

y=x+b

That is the equation with only the slope, however, we can use the point (-4,4) to find the y-intercept

4=-4+b

Add 4 both sides:

8=b

Thus the equation is y=x+8

Hope this helps!

3 0
2 years ago
Read 2 more answers
Find last​ year's salary​ if, after a 4 % pay​ raise, this​ year's salary is $ 33,280.
son4ous [18]

Answer: 31,948.80

Step-by-step explanation: I thin it his is correct. I multiplied the original amount by .04 then subtracted that from the original number.

6 0
2 years ago
Complete the simplified ratio:<br> 0.2 : 3.75
sesenic [268]

Answer:

1: 18.75

Step-by-step explanation:

We multiply both sides by 5.

4 0
3 years ago
Read 2 more answers
If
Leno4ka [110]

Answer:

\frac{s^2-25}{(s^2+25)^2}

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given:\mathcal{L}[t \cos 5t]=(-1)F'(s) with F(s)=\mathcal{L}[\cos 5t].

Now, F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt.

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s).

Solving for F(s) on the last equation, F(s)=\frac{s}{s^2+25}, then the Laplace transform we were searching is -F'(s)=\frac{s^2-25}{(s^2+25)^2}

3 0
3 years ago
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