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Alexxx [7]
3 years ago
13

Simplify: (8^2/3)^4

Mathematics
2 answers:
Keith_Richards [23]3 years ago
6 0

Answer:

Step-by-step explanation:

hello :

(8^2/3)^4 = ∛(8²) = ∛(4^3) = 4   because : 8² = 4^3 =64

Julli [10]3 years ago
5 0
8 8/3

Explanation:
(8 2/3)4
(8 2*4/3)
8 8/3
8 8/3=256=(8 2/3)4
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Order the decimals from least to greatest 48.6, 1.6, 0.13, 153.7, 2.2, 49.2
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0.13, 1.6, 2.2, 48.6, 49.2, 153.7

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6 0
3 years ago
A hardware store mixes paints in a ratio of three parts blue to five parts red to make two gallons of cabbage red . A ratio of s
Anna71 [15]
3b+5r=2 gallons of cabbage red
1r+7b=2gallons of egplant

1 gallon of cabbage sells for 22 (means 2 gallos is 44)
1 gallon of egplant is 26 (means 2 gallons is 52)

3b+5r=44
r+7b=52

r+7b=52
miinus 7b from both sides
r=52-7b
sub 52-7b for r in othe equestion
3b+5(52-7b)=44
3b+260-35b=44
-32b+260=44
minus 260 from both sides
-32b=-216
divide both sides by -32
b=6.75

r=52-7b
r=52-7(6.75)
r=52-47.25
r=4.75


3blue+5r=2gallons
5+3=8
8units=2gallons
 4units=1gallon
1unit=1quart

therefor

1 quart of blue paint is $6.75
1 quart of red paint is $4.75
5 0
4 years ago
A tank contains 250 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
Vilka [71]

Answer:

A(t)=250-230e^{-\frac{t}{50} }

Step-by-step explanation:

A(t) is the amount of salt in the tank at time t.

dA / dt = rate of salt flowing into the tank - rate of salt going out of the tank

dA / dt = (1 g/L)*(5 L/min) - (A(t)/250 g/L) * (5L/min)

dA / dt = 5 g/min - (A(t) / 50) g/min

\frac{dA}{dt}+\frac{A(t)}{50} = 5\\\\The\ integrating\ factor(IF)= e^{\int\limits \frac{1}{50}dt }=e^{\frac{t}{50} }\\\\Multiplying\ through\ by\ the\ I.F:\\\\\frac{dA}{dt}*e^{\frac{t}{50} }+\frac{A(t)}{50}*e^{\frac{t}{50} } = 5*e^{\frac{t}{50} }\\\\Integrating \ both \ sides:\\\\\int\limits[  \frac{dA}{dt}*e^{\frac{t}{50} }+\frac{A(t)}{50}*e^{\frac{t}{50} }] dt=\int\limits  5e^{\frac{t}{50} } dt\\\\A(t)e^{\frac{t}{50} } =\int\limits  5e^{\frac{t}{50} } dt\\\\

A(t)e^{\frac{t}{50} } =250e^{\frac{t}{50} }+C\\\\A(t)=250+Ce^{-\frac{t}{50} }\\\\But\ A(0) = 20\\\\20=250+Ce^{-\frac{0}{50} }\\\\C+250=20\\\\C=-230\\\\A(t)=250-230e^{-\frac{t}{50} }

7 0
3 years ago
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