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3 years ago

Please help me with my math homework??

Mathematics

1 answer:

qaws [65]3 years ago

4 0

-2 1/2 / 6 would be -5/12, just turn - 2 1/2 into a decimal and divide by 6 :)

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6c - (3 – 3c) = 24<br> Solve equation step by step

kap26 [50]

Hi there! :)

$\large\boxed{c = 3}$

6c - (3 - 3c) = 24

Distribute the negative sign with terms inside the parenthesis;

6c - (3) - (-3c) = 24

6c - 3 + 3c = 24

Combine like terms:

9c - 3 = 24

Add 3 to both sides:

9c = 27

Divide both sides by 9:

c = 3.

7 0

3 years ago

Read 2 more answers

Answer if you want to

prohojiy [21]

Answer:

B: 24(0.3)

Step-by-step explanation:

24 * 0.3 = 7.2

So Mandy spent $7.20 on arts and crafts.

24 - 7.2 = $16.80 remaining.

So this is the final answer is 24 * 0.3

5 0

3 years ago

What three numbers must be multiplied by 6

Advocard [28]

Answer:

6, 36, 60 have to be multiplied by six in order to get the answer.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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

Juan manages an appliance store. In July, sales were $161,800. In August, sales were $117,305. What was the rate of change? That

babunello [35]

I think it is a 27.5% decrease

4 0

3 years ago