Help with these please and highly appreciate the help !!!!!!!

Mathematics

1 answer:

k0ka [10]3 years ago

7 0

Which one will we start with? I'll explain 21. multiply it out and the clean it up. you should end up with 140=A if I'm correct.

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Will mark Brainliest!!!<br> Try to show a little work!❤️❤️❤️❤️❤️❤️❤️

Ymorist [56]

Answer:

Can you post 4.3? We need more information to answer this.

Step-by-step explanation:

4 0

3 years ago

To graph ( 0, 15 ) what is the only direction you will move

Dominik [7]

Answer:

you will only move 15 units up

Step-by-step explanation:

since the y is the only one that has a value, then you will have to focus more on the direction where the y will go. and since the y's value is positive then you will have to go upwards.

3 0

3 years ago

2 (-X + 3) + x <br><br> Ⓘ ⓡⓔⓐⓛⓛⓨ ⓓⓞⓝ'ⓣ ⓤⓝⓓⓔⓡⓢⓣⓐⓝⓓ

Darina [25.2K]

Answer:

-x +6

Step-by-step explanation:

We assume you want to simplify this.

Use the distributive property to eliminate the parentheses. That property tells you that the factor outside parentheses (2) will multiply both of the terms inside parentheses. It's as though you had a bag (parentheses) with two objects inside. Two such bags will have two of each of those objects.

2(-x +3) +x

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

= -2x +6 +x

Now, the like terms -2x and +x can be combined.

= x(-2 +1) +6

= x(-1) +6

= -x +6 . . . . . the simplified expression

5 0

3 years ago

Use theorem 7. 4. 2 to evaluate the given laplace transform. do not evaluate the convolution integral before transforming. (writ

irga5000 [103]

With convolution theorem the equation is proved.

According to the statement

we have given that the equation and we have to evaluate with the convolution theorem.

Then for this purpose, we know that the

A convolution integral is an integral that expresses the amount of overlap of one function as it is shifted over another function.

And the given equation is solved with this given integral.

So, According to this theorem the equation becomes the

$\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{\frac{s+1}{(s+1)^2+1}}{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).$