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Alex73 [517]
3 years ago
6

Marie has 50 pages in her notebook. She made a graph of the kinds of writing on all of the pages she has

Mathematics
1 answer:
MissTica3 years ago
4 0

Answer:

31

Step-by-step explanation:

50 - 2-3-6-3-5 = 31

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Find the area of the figure given.
kupik [55]
The answer is C - 63.2 cm
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3 years ago
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For what values of x and y must each figure be a parallelogram?
Illusion [34]

Answer:

x=11,y=10.

Step-by-step explanation:

The diagonals of a parallelogram bisect each other.

We can form an equation in x and y using this property.

2y+2=2x

and x+9=2y.

Solving the second  equation for x ,

x=2y-9.

Substituting x value in 2y+2=2x

2y+2=2(2y-9)

Or ,2y+2=4y-18.

Adding 18 both sides:

2y+20=4y.

Subtracting 2y both sides

20=2y.

Dividing both sides by 2:

y=10.

x=2y-9

x=2(10)-9 ( substituting y value)

x=20-9=11

x=11 and y=10 will make the figure a parallelogram.

4 0
3 years ago
Read 2 more answers
Two times Noura's age plus 5 years equals Ali's age. Ali is 19 years old. What formula could be used to calculate Noura's age
jenyasd209 [6]

Answer:

Noura is seven years old

Step-by-step explanation:

to get Noura's age first you need to subtract 5 years from Ali's age which will get you to 14 years of age now you just need to divide 14 by two to get Noura's age which is seven

3 0
3 years ago
What is the sum of 10/12and 11/12? Make sure to show all work to receive full credit!
sergiy2304 [10]

Answer:

1\frac{3}{4}

Step-by-step explanation:

\frac{10}{12} +\frac{11}{12} \\\\\frac{21}{12} \\\\1\frac{9}{12} \\\\1\frac{3}{4}

7 0
3 years ago
A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo
djverab [1.8K]

Answer:

a) y(t)=50000-49990e^{\frac{-2t}{25}}

b) 31690.7 g/L

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is \frac{dy}{dt}=R_{i}-R_{o}, where R_{i} is the rate of salt entering and R_{o} is the rate of salt going outside.

Then we have, R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}, and

R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}

So we obtain.  \frac{dy}{dt}=4000-\frac{2y}{25}, then

\frac{dy}{dt}+\frac{2y}{25}=4000, and using the integrating factor e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}, therefore  (\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}, we get   \frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}, after integrating both sides y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C, therefore y(t)= 50000 +Ce^{\frac{-2t}{25}}, to find C we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions y(0)=10, so 10= 50000+Ce^{\frac{-0*2}{25}}

10=50000+C\\C=10-50000=-49990

Finally we can write an expression for the amount of salt in the tank at any time t, it is y(t)=50000-49990e^{\frac{-2t}{25}}

b) The tank will overflow due Rin>Rout, at a rate of 80 L/min-40L/min=40L/min, due we have 500 L to overflow \frac{500L}{40L/min} =\frac{25}{2} min=t, so we can evualuate the expression of a) y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7, is the salt concentration when the tank overflows

4 0
3 years ago
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