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3 years ago

PLZ HELP!!!! I'll mark as brainly if you get it right

Mathematics

2 answers:

sasho [114]3 years ago

8 0

Answer:

$135.51

hope this is rught and hope this helps

choli [55]3 years ago

6 0

Answer:$ 137.51
2) 10.34 you can add up to at least 5 to 6 toppings.
Step-by-step explanation:
1) Subtract the two numbers
2)Add the first number to get 10.34, you can add up to at least 5 to 6 toppings.
Give me brainliest

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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 $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

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3 years ago

Solve the system of linear equations using the

MArishka [77]

Answer:

no solution

Step-by-step explanation:

y = -2x + 3

6x + 3y = -3

the substitution method means you plug one equation into the next, because the first equation gives us a solution for y we can go ahead and plug that into y of the second equation

6x + 3(-2x + 3) = -3

6x - 6x + 9 = -3

9 = -3

which is false meaning that there are no solutions and the lines don't touch at any point

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