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Delicious77 [7]
4 years ago
15

Gabe​ Amodeo, a nuclear​ physicist, needs 60 liters of a 60​% acid solution. He currently has a 50% solution and a 70% solution.

How many liters of each does he need to make the needed 60 liters of 60​% acid​ solution? Gabe needs 30 liters for a 50% solution .
He also needs a ___ for a 70% solution . whats __?
Mathematics
1 answer:
FromTheMoon [43]4 years ago
6 0

Answer:

Gabe needs 20 liters of the 40% solution.  

He also needs 40 liters of the 70% solution.

Step-by-step explanation:

Equation:


acid + acid = acid


0.40x + 0.70(60-x) = 0.60*60


40x + 70*60 - 70x = 60*60


-30x = -10*60


x = 20 liters


Gabe needs 20 liters of the 40% solution.  

He also needs 40 liters of the 70% solution.

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DiKsa [7]

9.

By the Segment Addition Postulate, SAP, we have

XY + YZ = XZ

so

YZ = XZ - XY = 5 cm - 2 cm = 3 cm

10.

M is the midpoint of XZ=5 cm so

XM = 5 cm / 2 =  2.5 cm

11.

XY + YM = XM

YM = XM - XY = 2.5 cm - 2 cm = 0.5 cm

12.

The midpoint is just the average of the coordinate A(-3,2), B(5,-4)

M = (A+B)/2 = \left( \dfrac{-3 + 5}{2}, \dfrac{2 + -4}{2} \right)

Answer: M is (1,-1)

You'll have to plot it yourself.

13.

For distances we calculate hypotenuses of a right triangle using the distnace formula or the Pythagorean Theorem.

AB^2 =(5 - -3)^2 + (-4 - 2)^2 = 8^2 + 6^2 = 100

Answer: AB=10

M is the midpoint of AB so

Answer: AM=MB=5

14.

B is the midpoint of AC.   We have A(-3,2), B(5,-4)

B = (A+C)/2

2B = A + C

C = 2B - A

C = ( 2(5) - -3, 2(-4) - 2 ) = (13, -10)

Check the midpoint of AC:

(A+C)/2 = ( (-3 + 13)/2, (2 + -10)/2 ) = (5, -4) = B, good

Answer: C is (13, -10)

Again I'll leave the plotting to you.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
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A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

and so (b) the amount of sugar in the tank at time t is

S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

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4 years ago
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