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4 years ago

5x + 20 =3x +66 please help me with this problem

Mathematics

2 answers:

andreev551 [17]4 years ago

5 0

5x+ 20 = 3x +66
-3x -3x
2x+20=66
-20 -20
2x=46
/2 /2
X=23

lakkis [162]4 years ago

4 0

First we need to combine like terms. so we move the like terms to different parts of the equation:
5x -3x = 66-20
then we combine the like terms:
2x= 46
bow we divide both sides by 2 to isolate x:
2x ÷2= 46 ÷2
x= 23
any questions? just ask. I hope this helped! :)

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A boat rental shop rents paddle boats for a fee plus an additional cost per hour. The cost of renting different numbers of hours

mylen [45]

The independent variable is that whose values do not take into account the values of other variables. That is the time in hours for this item. Then, for the dependent variable, the answer would be the cost of renting. The value of the dependent variable is based on the changes done in the values of the independent variable.

4 0

3 years ago

You have one cubic mile of water. you release it at 1000 gallons per

Tatiana [17]

It will take 764664.6875 days to empty the water

<h3>How to determine the time to empty?</h3>

The given parameters are:

Volume = 1 cubic mile
Rate = 1000 gallons per minutes

The time is calculated as:

Time = Volume/Rate

So, we have:

Time = 1 cubic mile ÷ 1000 gallons per minute

Convert gallons to cubic mile

Time = 1 cubic mile ÷ (9.0817e-13 * 1000) cubic mile per minute

Divide

Time = 1.10111715 × 10^9 minute

Convert to days

Time = 764664.6875 days

Hence, it will take 764664.6875 days to empty the water

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2 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

<u>Answer-</u>

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$