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lyudmila [28]
2 years ago
7

Glenda wrote 2/10 of her paper on Monday, 1/10 of her people on Tuesday and 1/10 of her paper on Wednesday . She says she will m

ore than half of her paper . Is she correct ? Why or why not ?
Mathematics
2 answers:
eimsori [14]2 years ago
6 0
If you are asking if she wrote half her paper:

Add the portions she has written.

= 2/10 + 1/10 + 1/10
= 4/10
simplify by 2
= 2/5 (or 0.4) has been written

2/5 < 1/2
0.4 < 0.5


ANSWER: She is not correct. She has not written 1/2 of her paper because 2/5 (or 0.4) is less than 1/2 (or 0.5).

Hope this helps! :)

Alecsey [184]2 years ago
3 0
Glenda is incorrect. shes wrong because she wrote 4/10 thats close to 1/2 but its not one half
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The commutative property states that you can rearrange the order of the numbers and get the same result. The commutative property states that you can rearrange the order of the numbers and get the same result. The addition properties are the exact same, but replace multiply with add. Your answer is D.

Step-by-step explanation:

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2 years ago
Help me plz. unit rate is 1.5 cups for every 1 minute
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6 0
3 years ago
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
aliya0001 [1]

Answer:

A=1500-1450e^{-\dfrac{t}{250}}

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

R_{in} =(concentration of salt in inflow)(input rate of brine)

=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}

R_{out}=(concentration of salt in outflow)(output rate of brine)

=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}

Now, the rate of change of the amount of salt in the tank

\dfrac{dA}{dt}=R_{in}-R_{out}

\dfrac{dA}{dt}=6-\dfrac{A}{250}

We solve the resulting differential equation by separation of variables.  

\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}

Taking the integral of both sides

\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}

Recall that when t=0, A(t)=50 (our initial condition)

50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}

4 0
3 years ago
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