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2 years ago

Who could help me with this ?

Mathematics

1 answer:

Ainat [17]2 years ago

7 0

Make sure to not use the useless numbers on the slant. Also remember to divide by two at the end of calculating.

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Which decimal is equal to three-fifths<br> 0.6 0.15 0.2 0.3 0.4

alexandr402 [8]

.6 is the decimal form of 3/5

6 0

2 years ago

Read 2 more answers

The perimeter of a square of side 5 cm is 25 cm.<br>O True<br>O Falsethe perimeter of a circle

larisa [96]

Answer:

false

Step-by-step explanation:

I would assume false as if each of the fours sides of the square are 5 cm, then doing the equation 5×4 (unsing 4 for the each of the congruent sides) would give up 20 cm.

4 0

2 years ago

2 ^( 3 - 9) - 11 how do I solve this

Setler [38]

2^(3-9)-11 =

1 step: subtract the 3 and 9 to get 6

(3-9)=6

2 step: multiply 2 to its self 6 times and
you should get 64

2*2=4*2=8*2=16*2=32*2=64

3 step: subtract 11 by 64 to get 53

64-11=53

hope I help if so hit the brainliest thanks

5 0

2 years ago

Express the given product as a sum or difference containing only sines or cosines.

beks73 [17]

Answer:

sin(8x)

Step-by-step explanation:

assuming that you didn't mean to write it twice

4 0

3 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