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MariettaO [177]
3 years ago
14

Franklin bought a book with 90 pages. If he

Mathematics
1 answer:
BartSMP [9]3 years ago
3 0
The answer is A) 60%
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Use the graph to evaluate the function.<br> f(-5) =<br> y<br> -8 -6
DENIUS [597]

Answer:

-4

Step-by-step explanation:

Go to x value of -5 and go down until you touch line.

3 0
3 years ago
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

7 0
3 years ago
Please I need help with this!!!!!!!
inysia [295]

Answer:

Step-by-step explanation:

1). Step 4:

   x=5^{\frac{4}{3}}=(5^4)^{\frac{1}{3}}

   x=\sqrt[3]{5^4} [Since, a^{\frac{1}{3}}=\sqrt[3]{a}]

   x=\sqrt[3]{5\times 5\times 5\times 5}

   Step 5:

   x=\sqrt[3]{(5)^3\times 5}

   x=\sqrt[3]{5^3}\times \sqrt[3]{5}

2). He simplified the expression by removing exponents from the given expression.

3). Let the radical equation is,

   (3x-1)^{\frac{1}{5}}=2

   Step 1:

   (3x-1)^{\frac{1}{5}\times \frac{5}{1} }=2^{\frac{5}{1}}

   Step 2:

   (3x-1)=2^5

   Step 3:

   3x=32+1

   Step 4:

   x=11

4). By substituting x=11 in the original equation.

   (3\times 11-1)^{\frac{1}{5}}=(32)^\frac{1}{5}

                         =(2^5)^\frac{1}{5}

                         =2

There is no extraneous solution.

3 0
2 years ago
A recipe says it produced the best sweet tea. In addition to using tea bags, the recipe calls for ½ cup of sugar for every 1 ½ c
Sergio [31]

Answer:

The relationship between the number of cups of water and sugar  in the recipe is w = ¹/₃(s)

Step-by-step explanation:

Given;

number of cups of sugar = s

number of cups of water = w

The relationship between the number of cups of water and sugar in this recipe is given as;

½ cup of sugar is proportional to 1 ½ cups of water

\frac{1}{2} s = \frac{3}{2}w\\\\s= 3w\\\\w = \frac{s}{3}

Therefore, the relationship between the number of cups of water and sugar  in the recipe is w = ¹/₃(s)

7 0
2 years ago
Read 2 more answers
Please help me please ! Quick
UkoKoshka [18]

Answer:

4,2 is tge answer due to the interaction point plus it says the solution is at the interaction.

8 0
2 years ago
Read 2 more answers
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