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2 years ago

Help me please!!! It’s due at 11:59

Mathematics

1 answer:

Temka [501]2 years ago

7 0

Answer:

Step-by-step explanation:

Let the required number be x.

write the expression for the difference of the number and 4.

=> x - 4

the question equates 5 times the difference with some other terms, so find the equation for 5 times the difference.

=> 5( x - 4)

The equation in Step 2 is said to be equal to "two less than four times the number".

Find the expression for four times the number

=> 4x

Find the expression for two less than 4x.

=> 4x - 2

Equate the terms that it says are equal:

=> 5(x - 4) = 4x - 2

distributing 5 within the parenthesis:

=> 5x - 20 = 4x - 2

taking like terms on one side:

=> 5x - 4x = 20 - 2

(the sign carried by the terms get inverted while changing their sides)

=> x = 18

and x was the required number, so the answer Is 18. 

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The product of 4 and a number f decreased by 2 is 15

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Answer: The number is: "2 " .

__________________________________________

Step-by-step explanation:

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Write the expression; which is an equation, as follows:

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" 4x − 12 = 2(-x) " ; in which "x" represents "the number for which we shall solve" .

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If the "number" = "x" ; the "opposite of the number" = " -x " ;

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Rewrite as: " 4x − 12 = -2x " ;

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Answer: The number is: "2 " .

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Let us check our answer, by plugging in "2" for "x" in our original equation:

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→ " 4x − 12 = 2(-x) " ;

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3 years ago

Read 2 more answers

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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:

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

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

The instantaneous amount of salt in the tank is:

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