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3 years ago

Can you plz help me with this question thx :)

Mathematics

1 answer:

aalyn [17]3 years ago

7 0

Answer:

where is r

if you reply maybe you get answer in com.ents

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Whats f? 0.62(f + –6) + 6.96 = 8.2

Semmy [17]

Answer:

f=8

Step-by-step explanation:

0.62(f+(-6)+6.96=8.2

0.62(f-6)+6.96=8.2

multiply both sides by 100

0.62(f-6)*100+6.96*100=8.2*100

Refine

62(f-6)+696=820

Subtract 696 from both sides

62(f-6)+696-696=820-696

Simplify

62(f-6)=124

Divide both sides by 62

62(f-6)/62 = 124/62

Simplify

f-6=2

add 6 to both sides

f-6+6=2+6

simplify

f=8

5 0

2 years ago

Read 2 more answers

I need to know the answer quickly.

alisha [4.7K]

4,7, second bubble for the word part

5 0

2 years ago

What is 1/100th of 882

Lostsunrise [7]

The answer is 8.82 because if you divide 882 by 100 it gives you 8.82

5 0

3 years ago

Read 2 more answers

I NEED THE ANSWER ASAPPP!!! help please

Sidana [21]

Answer:

Step-by-step explanation:

x^2 + 8^2 = 9^2

x^2 + 64 = 81

x^2 = 17

x = sqrt(17)

answer is D

7 0

3 years ago

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago