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tangare [24]
3 years ago
6

A store has 125 scientific calculators.

Mathematics
1 answer:
aleksley [76]3 years ago
7 0

Answer:

Brand x: 80

Brand y: 45

Step-by-step explanation:

x + y = 125

20x + 15y = 2275

    -20x - 20y = -2500

(+)  20x + 15y = 2275

--------------------------------

               -5y = -225

y = 45

x + y = 125

x + 45 = 125

x = 80

Answer:

Brand x: 80

Brand y: 45

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Answer:

m∠1=103°

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

An exterior angle of a triangle is equal to the sum of the opposite interior angles

so

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m∠1=m∠2+m∠3

substitute the given values

m∠1=31°+72°=103°

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5. Which graph shows a line with the same y-intercept as the graph of 10x-16y=-40?
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Step-by-step explanation:

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What is the area of this parallelogram?<br><br><br> 104 m²<br><br> 88 m²<br><br> 56 m²<br><br> 32 m²
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6 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
Dx+hy=j,forx   x=    thanx for any and all help
Crazy boy [7]
dx+hy=j \\ \\ dx = j-hy \ \ / : \ d \\ \\d\neq 0\\ \\x=\frac{j-hy }{d}


6 0
3 years ago
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