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ryzh [129]
3 years ago
10

Find the area of a triangle with legs that are: 6 mm, 4 mm, and 8 mm.

Mathematics
1 answer:
Musya8 [376]3 years ago
6 0
Hence,the option is B. If you have any doubts plzz lemme know.

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Please help me with the below question.
Snezhnost [94]

6a. By the convolution theorem,

L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}

6b. Similarly,

L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}

7. Take the Laplace transform of both sides, noting that the integral is the convolution of e^t and f(t).

\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau

\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)

where g(t) = e^t. Then G(s) = \frac1{s-1}, and

F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}

We have the partial fraction decomposition,

\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)

Then we can easily compute the inverse transform to solve for f(t) :

F(s) = -\dfrac1s + \dfrac4{s+3}

\implies \boxed{f(t) = -1 + 4e^{-3t}}

6 0
2 years ago
Which single transformation can be applied to the graph of y = 72 StartRoot x EndRoot to produce the graph of y = 9 StartRoot x
Ber [7]

Answer:

vertical....

Step-by-step explanation:

NawfSide 38 Baby

3 0
3 years ago
Simplify (-2 + 3i) + 2(5 + 6i). Enter your answer in the form a + bi.
Reptile [31]

We will have the following:

(-2+3i)+2(5+6i)=-2+3i+10+12i

=(10-2)+(3i+12i)=8+15i

So, the solution would be:

8+15i

3 0
1 year ago
What is the total of 4/7 +3/7
lilavasa [31]

Answer:

7/7 also known as a whole (1)

Step-by-step explanation:

All you have to do is add 4 and 3 then keep the denominator the same (7)

6 0
3 years ago
There are 32 peanuts in a bag. Elliot takes 25% of the peanuts from the bag . Then Zaire takes 50% of the remaining peanuts.How
salantis [7]
There wouldbe 25% left  i believe
6 0
4 years ago
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