All categories

2 years ago

Can someone plz help this is like my 5th time putting up this question!!!

Mathematics

1 answer:

serg [7]2 years ago

3 0

Answer:

the table should be 25,98,173,248

You might be interested in

Please help me click the picture to see

artcher [175]

0.25x+97
Factoring:
Convert 0.25 to a fraction:
1/4x+97
Factor out 1/4:
Solution:
1/4(x+388) or 0.25(x+388)

6 0

2 years ago

The system of equations represented by lines p and q is shown on the graph below. Based on the graph, what is the solution of th

exis [7]

Answer:

C(2,1)

Step-by-step explanation:

8 0

2 years ago

Raymoon sold an article at a loss of 5%. If he had sold it for 270 more, he would have gained 7% find the cost prize of article

mote1985 [20]

IᑎᔕEᑕTIOᑎ ᑭEᒍᗩY✓ hahahaha

7 0

2 years ago

To the nearest hundredth, what is the value of x? <br> 36.08<br> 41.51<br> 47.81<br> 72.88

Crazy boy [7]

Answer:

Step-by-step explanation:

Using the law of sines, we can make a proportion.

But first, we'll need to solve for the unknown angle.

We add up the two known angles and subtract that by 180.

90 + 41 = 131

180 - 131 = 49

So the unknown angles is 49.

Then, we can use the law of sines.

Make the equation.

sin(90)/55 = sin(49)/x

Simplify this using a calculator and you get around 41.51 or option B.

5 0

2 years ago

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

2 years ago