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2 years ago

1. Is c=71 a solution to the equation 56 = c - 15 ?

Mathematics

1 answer:

jolli1 [7]2 years ago

7 0

Answer:

1. Yes

2. Yes

3. No

Step-by-step explanation:

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What is the system equation of y=1/4x : x+4y=8?

Masja [62]

Since x+4y = 8 the thing is y = 1/4*x but if you get 1/4 of x then multiply it by 4 when you put it back in the problem, which means x=4 y=1

3 0

3 years ago

Can someone help on question 2

Kobotan [32]

Step-by-step explanation:

i think option a is the correct answer

5 0

3 years ago

Read 2 more answers

Find the inverse laplace<br><br> F(s)=11s/ s^2-12s+52

lions [1.4K]

Answer:

$11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t$

Step-by-step explanation:

We can write $\frac{11s}{s^2-12s+52}$ as follows:

$\frac{11s}{s^2-12s+52}\\=11\left [ \frac{s}{s^2-12s+52} \right ]\\=11\left [ \frac{s}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6+6}{(s-6)^2+16} \right ]\\=11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16}$

To find:

$L^{-1}\left [ \frac{11s}{s^2-12s+52 \right ]}\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]$

We will use formulae:

$L^{-1}\left \{ \frac{s-a}{(s-a)^2+b^2} \right \}=e^{at}\cos bt\\L^{-1}\left \{ \frac{b}{(s-a)^2+b^2} \right \}=e^{at}\sin bt$

we get solution as :

$L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+16} \right ]+\frac{66}{(s-6)^2+16} \right ]\\=L^{-1}\left [ 11\left [ \frac{s-6}{(s-6)^2+4^2} \right ]+\frac{66}{4}\left [ \frac{4}{(s-6)^2+4^2} \right ] \right ]\\=11e^{6t}\cos 4t+\frac{33}{2}e^{6t}\sin 4t$

5 0

3 years ago

an airplane travels 4688 kilometers against the wind in 8 hours and 5808 kilometers with the wind in the same amount of time. wh

Vladimir [108]

The speed of wind is 656 kmph and speed of plane is 70 kmph.

<u>SOLUTION: </u>

Given that, an air-plane travels 4688 kilometers against the wind in 8 hours

And 5808 kilometers with the wind in the same amount of time.

We have to find the rate of the plane in still air and the rate of wind

Now, let the speed of wind be a kmph and speed of plane be b kmph.

And we know that, $\text { distance }=\text { speed } \times \text { time }$

$\begin{array}{l}{\text { Then, while travelling with wind } \rightarrow 5808=(a+b) \times 8 \rightarrow a+b=726 \rightarrow a=726-b \rightarrow(1)} \\\\ {\text { While travelling against wind } \rightarrow 4688=(a-b) \times 8 \rightarrow a-b=586 \rightarrow(2)}\end{array}$