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2 years ago

PLEASE HELP GEOMETRY !!!!!

Mathematics

1 answer:

statuscvo [17]2 years ago

8 0

Answer:

x = 11.5

Step-by-step explanation:

$\sin \: 35 \degree = \frac{x}{20} \\ \\ x = 20 \sin \: 35 \degree \\ \\ x = 11.471528727 \\ \\ x = 11.5$

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2 years ago

Write the sentence as an equation.<br> 62 more than the product of 150 and r is 40

Elenna [48]

Answer:

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Step-by-step explanation:

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2 years ago

Identify the expression equivalent to 4x + x + 7) - 2x+8 - 4 by substituting x= 1 and x = 2. А. 6x + 11 B. 3(x+7) C. 2(3x + 16)

Alex

Answer:

4x + x + 7) - 2x+8 - 4 by substituting x= 1 and x = 2. А. 6x + 11 B. 3(x+7) C. 2(3x + 16) D. 3x + 16

Step-by-step explanation:

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2 years ago

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 :

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

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2 years ago

Order of operations

Marat540 [252]

Answer: d^(2) + (ab)/(c)

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers