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nignag [31]
3 years ago
12

Leezas aquarium holds 55 gallons of water. She is filling the tank and has already put in 22 gallons. Write and solve an inequal

ity to determine out how many more gallons she might put in the tank.
Mathematics
1 answer:
svp [43]3 years ago
6 0

Leeza can put 33 or less gallons of water in aquarium.

Step-by-step explanation:

Given,

Capacity of aquarium = 55 gallons of water

Water already in aquarium = 22 gallons

Let,

x be the number of gallons that can be added in aquarium.

As aquarium can hold 55 gallons of water, therefore, we cannot add more than 55 gallons of water.

x+22\leq 55

Subtracting 22 from both sides

x+22-22\leq 55-22\\x\leq 33

Leeza can put 33 or less gallons of water in aquarium.

Keywords: inequality, subtraction

Learn more about inequalities at:

  • brainly.com/question/7294502
  • brainly.com/question/7490805

#LearnwithBrainly

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Answer:

2m+2=16, then do the isolate m but doing the inverse operation of 2 which is now -2 and add that to 16 would be left with 2m=14 and 14/2 is 7 so m=7

Step-by-step explanation:

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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

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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}

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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

brainly.com/question/14487937

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2 years ago
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dolphi86 [110]

Answer:

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Step-by-step explanation:

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4 5

(Simplify)

4 ÷ 2

4 5

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20 ÷ 8

20 20

(Divide)

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3 years ago
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