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olchik [2.2K]
3 years ago
6

Find the missing length

Mathematics
1 answer:
wel3 years ago
6 0

Answer:

I think it is 35

Step-by-step explanation:

Proportions

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F(x) = x ^ 2 + 6x - 4 in vertex form
Svetach [21]

Answer:y=(x-6)^2-4

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Explain what i did in a detailed answer.
forsale [732]

Answer:

Step-by-step explanation: You added the following numbers multiplied the product by 5 = 17 and then you abbrieviated it and reapeated the process

3 0
3 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
Find the inverse of the function f(x) = (x - 4) 2 - 5 if x ≥ 4.
Serggg [28]
Same here, we do a quick switcharoo on the variables first,

\bf \stackrel{f(x)}{y}=(x-4)^2-5\qquad inverse\implies \boxed{x}=\left( \boxed{y}-4 \right)^2-5
\\\\\\
x+5=(y-4)^2\implies \pm\sqrt{x+5}=y-4\implies \pm\sqrt{x+5}+4=y
6 0
3 years ago
Solve the equation.
Vinil7 [7]
If you would like to solve the equation x + 1/6 = 6, you can do this using the following steps:

x + 1/6 = 6       /-1/6
x + 1/6 - 1/6 = 6 - 1/6
x = 6 - 1/6
x = 36/6 - 1/6
x = 35/6
x = 5 5/6

The correct result would be C. x = 5 5/6.
5 0
3 years ago
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