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nadezda [96]
3 years ago
13

-1=z\3-7 what's the answer

Mathematics
2 answers:
baherus [9]3 years ago
7 0
Z=-4 ......................
valentina_108 [34]3 years ago
5 0
Z=-4 !~!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Simplify: 64 ∙ 42 − 3 ∙ 22
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59070

Step-by-step explanation:

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A printer takes 30 minutes to print a banner and a poster. I will take 50 minutes to print a banner and three posters. How long
qwelly [4]

Answer:

It takes the printer 20 minutes to print a banner

Step-by-step explanation:

Notice that there are two unknowns in this problem: the time to print a poster and the time to print a banner. Let's assign the letter "b" to the time it takes to print a banner, and "p' the time it takes to print a poster.

So we can write the following two equations involving the total amount of time that takes to print in the two cases:

b+p=30\\b+3p=50

We can solve for the unknown "p" by subtracting term by term in the equation set above:

b+3p=50\\b+p=30\\\\b-b+3p-p=50-30\\2p=20\\p=10

So it takes 10 minutes to print each poster, we can find the time it takes to print the banner by using the value we just found in one of the equations:

b+p=30\\b+10=30\\b=20

So, it takes 20 minutes to print a banner.

8 0
3 years ago
Initially 5 grams of salt are dissolved into 10 liters of water. Brine with concentration of salt 5 grams per liter is added at
DiKsa [7]

Salt flows in at a rate of (5 g/L)*(3 L/min) = 15 g/min.

Salt flows out at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.

So the net flow rate of salt, given by x(t) in grams, is governed by the differential equation,

x'(t)=15-\dfrac{3x(t)}{10}

which is linear. Move the x term to the right side, then multiply both sides by e^{3t/10}:

e^{3t/10}x'+\dfrac{3e^{t/10}}{10}x=15e^{3t/10}

\implies\left(e^{3t/10}x\right)'=15e^{3t/10}

Integrate both sides, then solve for x:

e^{3t/10}x=50e^{3t/10}+C

\implies x(t)=50+Ce^{-3t/10}

Since the tank starts with 5 g of salt at time t=0, we have

5=50+C\implies C=-45

\implies\boxed{x(t)=50-45e^{-3t/10}}

The time it takes for the tank to hold 20 g of salt is t such that

20=50-45e^{-3t/10}\implies t=\dfrac{20}3\ln\dfrac32\approx2.7031\,\mathrm{min}

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