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Oxana [17]
3 years ago
6

Can anyone point me in the direction of a good precalc study material or cheat sheet?​

Mathematics
1 answer:
finlep [7]3 years ago
6 0

Answer:

patrickJMT

Brian McLogan

Step-by-step explanation:

Really good teachers, I get most of my pre-calc help from them. You can also just ask some questions on brainly if you get stuck.

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d1i1m1o1n [39]
2. So for this question, you do the distributive property, which is multiplying the term outside of the parenthesis by the terms in the parenthesis. In this case, the outside term is 1/2. 8x * 1/2 = 4x and 1/2 * -16 is -8. That brings us to 4x - 8, which is our answer. Therefore, the correct answer choice is B.

3. Okay. In this problem, we are talking about the sum of 12 and a number. That's 12 + n, because sum means the answer to the addition problem D is out, because we don't do distributive property in this case. The sum of 12 + n is 4n. The expression is 12 + n = 4n. The answer is B.
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3 years ago
A frozen yogurt shop offers 10 different flavors of yogurt and 15 different toppings. How many double-dip sundae combinations ca
Aleksandr [31]
25 yogurt dips 
i took this quiz and that was the answer 


3 0
3 years ago
Read 2 more answers
What is the Absolute value 18/8
vlada-n [284]
18/8
8/8

1/8 = 0.125

1+0.125=1.125
6 0
3 years ago
The average annual salary of the employees of a company in the year 2005 was $80,000. It increased by the same factor each year
Kryger [21]
The answer is f(x)=80(1.1)x
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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