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2 years ago

14

Please help there are a few questions but I will give brainliest to first person who answers all of them! Thank you!!!!!!!!!!!!!

!!!!!!!!!!!

1 answer:

viva [34]2 years ago

7 0

Step-by-step explanation:

1. 57 degrees

2. B

3. 117 degrees

4. C

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Darya [45]

Answer:

x = 16.5

Step-by-step explanation:

Use pythagorean theorem

a² + b² = c²

Plug in terms

7.1² + x² = 18²

Square each term

50.41 + x² = 324

Subtract a from c

x² = 273.59

Find square root

16.5

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2 years ago

the athletic department spent $11,250 on the 15 sports at the school. what is the rate that the athletic department spent in eac

Sholpan [36]

Answer:

$11,250 divided by 15 is 750, which means that the answer is the rate is $750 dollars spent in each sport.

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2 years ago

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Answer pls!! thank u

mestny [16]

Answer: See below

Step-by-step explanation:

A linear line is a straight line. It can be going up, down, vertical, or horizontal. As long as it's a straight line, then it is a linear equation. Graph 1 and 5 are linear.

A quadratic graph is a parabola. It is in the shape of a "U". The "U" can be doing up or down. It can also be more than a parabola. Any x to th epower of a number is also a quadratic graph. Graph 2 and 4 are quadratic graphs.

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2 years ago

Please help me!!!!!””

Advocard [28]

Answer:

d=45t

Step-by-step explanation:

to find speed you divide the distance by the time. 67.5/1 1/2 = 45

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2 years ago

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15.30 find the inverse laplace transform of: 1. (a) f1(s) = 6s 2 8s 3 s(s 2 2s 5) 2. (b) f2(s) = s 2 5s 6 (s 1) 2 (s 4) 3. (c) f

EleoNora [17]

The solution of the inverse Laplace transforms is mathematically given as

$f_{1}(t)=e^{-t}\sin (2 t)$

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$

<h3>What is the inverse Laplace transform?</h3>

1)

Generally, the equation for the function is mathematically given as

$F_{1}(s)=\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}$

By Applying the Partial fractions method

$\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}=\frac{A}{s}+\frac{B s+C}{s^{2}+2 s+5}$

$6 s^{2}+8 s+3=A\left(s^{2}+2 s+5\right)+(B s+C) s$

$\begin{aligned}&3=5 A \\&A=\frac{3}{5}\end{aligned}$

Considers s^2 coefficient

$\begin{aligned}&6=A+B \\&B=6 \cdot A \\&B=\frac{27}{5}\end{aligned}$

Consider s coeffici ent

$\begin{aligned}&8=2 A+C \\&C=8-2 A \\&C=\frac{34}{5}\end{aligned}$

Putting these values into the previous equation

$&F_{1}(s)=\frac{3}{5 s}+\frac{27 s+34}{5\left(s^{2}+2 s+5\right)} \\\\&F_{1}(s)=\frac{3}{5 s}+\frac{27(s+1)}{5\left((s+1)^{2}+4\right)}+\frac{7 \times 2}{10\left((s+1)^{2}+4\right)}$

By taking Inverse Laplace Transforms

$f_{1}(t)=\frac{3}{5}+\frac{27}{5} e^{-t} \cos (2t) + \frac{7}{10}\\\\$

$f_{1}(t)=e^{-t}\sin (2 t)$

For B

$F_{2}(s)=\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}$

By Applying Partial fractions method

$\begin{aligned}&\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\\\&s^{2}+5 s+6=A(s+1)(s+4)+B(s+4)+C(s+1)^{2}\end{aligned}$

at s=-1

$1-5+6=3 B \\\\B=\frac{2}{3}$

at s=-4

$&16-20+6=9 C \\\\&9 C=2 \\\\&C=\frac{2}{9}$

at s^2 coefficient

1=A+C

A=1-C

A=7/9

inputting Variables into the Previous Equation

$\begin{aligned}&F_{2}(s)=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\&F_{2}(s)=\frac{7}{9(s+1)}+\frac{2}{3(s+1)^{2}}+\frac{2}{9(s+4)}\end{aligned}$

By taking Inverse Laplace Transforms

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

For C

$F_{3}(s)=\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}$

Using the strategy of Partial Fractions

$\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}=\frac{A}{s+1}+\frac{B s+C}{s^{2}+4 s+8}$

$10=A\left(s^{2}+4 s+8\right)+(B s+C)(s+1)$

S=-1

10=(1-4+8) A

A=10/5

A=2

Consider constants

10=8 A+C

C=10-8 A

C=10-16

C=-6

Considers s^2 coefficient

0=A+B

B=-A

B=-2

inputting Variables into the Previous Equation

$&F_{3}(s)=\frac{2}{s+1}+\frac{-2 s-6}{\left((s+2)^{2}+4\right)} \\\\&F_{3}(s)=\frac{2}{s+1}-\frac{2(s+2)}{\left((s+2)^{2}+4\right)}-\frac{2}{\left((s+2)^{2}+4\right)}$

Inverse Laplace Transforms

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$

Read more about Laplace Transforms

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3 0

1 year ago