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ivolga24 [154]
3 years ago
8

a sixth grade student can complete 20 multiplication facts in one minute if you works at the same rate how long should it take h

im to complete 35 multiplication facts? [type answer as decimal number.]​
Mathematics
2 answers:
Thepotemich [5.8K]3 years ago
8 0

Answer:

The answer is 1.75 minutes :)

Step-by-step explanation:

I used my calculator to figure this one out.

melamori03 [73]3 years ago
6 0

Answer:

It should take him <u>1.75 minutes</u> to complete 20 multiplication facts, which is the same thing as 1 minute and 45 seconds, or 105 seconds.

Step-by-step explanation:

a. If it takes him one minute to complete 20 multiplication facts, that's 3 seconds per question, because 60/20 = 3.

b. Multiply 3 x 35 questions to get 105 seconds.

c. Divide 105 seconds by 60 seconds (60 in a minute) to get 1.75 minutes, which is the same thing as 105 seconds.

Hope this helps! Feel free to give me Brainliest if you feel this helped. Have a good day and good luck on your assignment. :)

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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
If (42)^p = 41^4, what is the value of p?
babymother [125]

First you need to make both bases the same:

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To make the base of 42 equal to 41, you would have 41^x = 41

X - ln(42) / ln(41) = 1.00648904


Now you have 41^1.00648904(p) = 41^4


Now the bases are equal so we need to set the exponents to equal:

1.00648904(p) = 4

Divide both sides by 1.00648904 to solve for P

P = 4 / 1.00648904

P = 3.97421114

4 0
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6 0
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arlik [135]
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6x-4 = 2(3x-2)

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2 years ago
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