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3 years ago

Score: 7.5/50

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

6 0

Complete Question:

Annette and Rose went to an orchard to pick pears. Annette picked $8\frac{1}{6}$

pounds of pears and Rose picked $6\frac{1}{8}$ pounds. How much more did Annette pick than Rose? Write your answer as a reduced mixed number.

Answer:

Annette picked $2\frac{1}{24}$ pounds of pears more than Rose

Step-by-step explanation:

Pounds of pears picked by Annette = $8\frac{1}{6}$

Pounds of pears picked by Rose = $6\frac{1}{8}$

To get the difference in the pounds of pears picked by Annette and Rose, we will do a simple subtraction:

Difference in the pounds of pears picked = $8\frac{1}{6} - 6\frac{1}{8}$

$= 2(\frac{1}{6} - \frac{1}{8})\\= 2\frac{4 - 3}{24} \\= 2\frac{1}{24}$

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Answer:

In order to find the value of s, we must isolate the s variable in this equation, so the first step should be subtracting 3 5/6 from both sides of this equation. In order to subtract, we multiply each denominator to get 42, and multiply the numerator by the same amount, and subtract the number from both sides:

s + 3 35/42 = 9 6/42

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Step-by-step explanation:

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2 years ago

What is f(5) if f(1) = 3.2 and f(x + 1) = (f(x))?

scoray [572]

Answer:

The value of f(5) is 3.2.

Step-by-step explanation:

It is given that f(1) = 3.2 and

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Substitute x=1 in the above equation.

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Substitute x=2 in the given equation.

$f(2+1)=f(2)$

$f(3)=3.2$

Substitute x=3 in the given equation.

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Therefore the value of f(5) is 3.2.

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3 years ago

Read 2 more answers

A spaceship traveled 32 million miles. In scientific notation that is _____ miles. 3.2 × 10-7 3.2 × 10-6 3.2 × 106 3.2 × 107

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In scientific notation :
32 million miles is 3.2 x 10 ^7 miles.

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2 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

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Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
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Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

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$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

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2 years ago

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mote1985 [20]

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Step-by-step explanation:

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