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2 years ago

Elena is buying food for a school picnic and has $40.00 to spend on sandwiches.

Mathematics

1 answer:

ikadub [295]2 years ago

3 0

Answer:is 13 ok check and try

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What is 2 16/99 as a decimal

Serggg [28]

2 16/99 as decimal should be:
2 + (16/99) = 2 +0.17 = 2.17

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3 years ago

Which of the following formulas expresses h in terms of l and s?

STALIN [3.7K]

Answer: D

Step-by-step explanation:

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2 years ago

PLEASE HELP SOON! ALSO INCLUDE A SMALL EXPLANATION as well as the answer ! I am rewarding 100points if you answer quickly

alina1380 [7]

Answer:

the answer is B

f(x)=-2x+3

Step-by-step explanation:

to get the -2x I did --->

pick two points I used (0,3) and (1,1) and used the equation $\frac{y2-y1}{x2-x1}$

where y2 would be 1 and y1 would be 3

and where x2 would be 1 and x1 would be 0

so you would put the numbers in the equation which would look like $\frac{1-3}{1-0}=\frac{-2}{1}$ and equal -2 which would be -2x in the answer

And the +3 is the y-intercept

4 0

2 years ago

Read 2 more answers

Using π = 3.14, find the area of a circle with a radius of 4.2. Round your answer to the nearest hundredth.

denpristay [2]

Answer:

55.42

Step-by-step explanation:

The formula is A= pir^2

hope this helps

4 0

2 years ago

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>

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

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