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ryzh [129]
2 years ago
10

Which trinomials are perfect square trinomials?

Mathematics
2 answers:
telo118 [61]2 years ago
4 0
The correct answer is y2+20y+200
slavikrds [6]2 years ago
3 0
Answer is the following y2+25y+200
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I need help i don't get it !!!!!
joja [24]

Answer:

4

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

sin theta = opp / hyp

sin 30 = x/8

8 sin 30 =x

8 (1/2) = x

4

4 0
3 years ago
Read 2 more answers
Find the value of x and y
slava [35]

Answer:

Step-by-step explanation:

OS ≅ OU, so we will set those 2 expressions equal to each other and solve for y:

6y = 42 so

y = 7

Same goes for OT and OV:

x + 5 = 23 so

x = 18 and

SU = 84

5 0
3 years ago
If f(x) = integral of 1/ (sqrt of t^3 +2)dt, then f'1 = ?
mihalych1998 [28]
Differentiating an integral removes the integral.

f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
4 0
3 years ago
Marcus changed jobs after college. His old salary was $48000 per year. Now his new salary is 37% more per year. What is his new
spayn [35]

Answer:

65,760

Step-by-step explanation:

move decimal over two places to make a percent a decimal so .37

then multiply that times the 48000 to get 17,760 and then add that to the original 48000 to get 65,760

8 0
2 years ago
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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