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bija089 [108]
3 years ago
13

John can create an app in 6 weeks while Ben can do the same job in 12 weeks. How many weeks would it take them to create an app

together?
Mathematics
2 answers:
Oksana_A [137]3 years ago
8 0
6weeks cause John speeds up bens work
aleksley [76]3 years ago
7 0
6 because if thet are doing it togather jhon will speed things up
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A rectangular swimming pool that is 10 ft wide by 16 ft long is surrounded by a cement sidewalk of uniform width. If the area of
Alchen [17]
We are asked to solve for the width "x" in the given problem. To visualize the problem, see attached drawing.
We have the area of the swimming pool such as:
Area SP = l x w
Area SP = 10 * 16
Area SP = 160 feet2
Area of the swimming pool plus the sidewalk with uniform width:
Area SP + SW = (10 + x) * (16 + x)
160 + 155 = 160 + 10x + 16x + x2
160 -160 + 155 = 26x + x2
155 = 26 x + x2
x2 + 26x -155= 0
Solving for x, we need to use quadratic formula and the answer is 5 feet.

The value of x is <span>5 feet. </span>

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3 years ago
Whats the word phrase of 49+m?
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The sum of 49 and m.
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What is the value of s?
beks73 [17]

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To what

Step-by-step explanation:

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I have attached the question below
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2 years ago
(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi
Drupady [299]

Answer:

a. \dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

b. x(t) = 6\frac{2}{3} \cdot c \cdot t

c. c  = \dfrac{3}{8}  \ g/L

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx =  (5·c - 5·c/3)×dt = 20/3·c = 6\frac{2}{3} \cdot c \cdot dt

\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

b. The amount of salt, x after t minutes is given by the relation

\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

dx = 6\frac{2}{3} \cdot c \cdot dt

x(t) = \int\limits \, dx  = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt

x(t) = 6\frac{2}{3} \cdot c \cdot t

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10

6\frac{2}{3} \times c  =\dfrac{25 \ grams }{10}

c  =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} }  = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200}  \ g/L = \dfrac{3}{8}  \ g/L

c  = \dfrac{3}{8}  \ g/L

4 0
3 years ago
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