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3 years ago

15

jannie has 2 liters of orange squash she also has some empty cups One full Beach club holds 30 mL of orange squash jennine files

as many cups as possible How much orange squash does janine have after filing as many cups as possible

1 answer:

navik [9.2K]3 years ago

4 0

Answer:

The question is a bit incorrect as there isnt mentioned how many cups are there.But if you meant how many 30ml cups can Jannie fill it would be 66.66 cups or 67/66 cups .

Step-by-step explanation:

If you would correct your answer I would try my best to give you the brainliest answer.

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A farmer plants wheat and corn. It costs about $150 per acre to plant wheat and about $350 per acre to plant corn. The farmer pl

spayn [35]

The inequality that models this situation is 150x + 350 y ≤ 250000 and x + y < 1200

<h3>What is an inequality?</h3>

An inequality is an expression that shows the non equal comparison of two or more numbers and variables.

Let x represent the number of wheat acres and y represent the number of corn acres, hence:

150x + 350 y ≤ 250000 (1)

Also:

x + y < 1200 (2)

The inequality that models this situation is 150x + 350 y ≤ 250000 and x + y < 1200

Find out more on inequality at: brainly.com/question/24372553

3 0

3 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>

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

2 years ago

2. Write an expression for the perimeter of a rectangle with a length of

solmaris [256]

Answer:

<h2><u><em>2 × [(3x² + x + 2) + (-x² – 5x + 1)] =</em></u></h2>

Step-by-step explanation:

Write an expression for the perimeter of a rectangle with a length of

(3x² + x + 2) and a width of (-x² – 5x + 1).

P = 2 × (L + W)

<u><em>2 × [(3x² + x + 2) + (-x² – 5x + 1)] =</em></u>

if you want the result:

4x²−8x+6

3 0

1 year ago

New York City and New Jersey are approximately 90 miles apart. If a map is drawn using a scale of 2 cm = 15 miles, how far apart

defon

Answer:

12 cm

Step-by-step explanation:

6 cm because 90 divided by 15 is 6 and 6 times 2 is 12

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3 years ago

The diagrams below represent rock layers found in two different locations by two different scientists. The scientist at location

scZoUnD [109]

Answer:

Layers P, Q, R, and S are the same as layers IV, V, VI and VII.

8 0

3 years ago

Read 2 more answers