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3 years ago

Which shows how the distributive property can be used to evaluate 7×84/5

Mathematics

1 answer:

finlep [7]3 years ago

6 0

Answer:

Number 4

Step-by-step explanation:

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Which number line shows the correct placement of 2.4 to the second power

MakcuM [25]

Where are the number line choices!?

6 0

3 years ago

1. Determine the percent of all voters that did not vote Republican or Democrat but voted "other" instead. Round your answer to

lyudmila [28]

Answer:

1. 10%

2. 49.1%

Step-by-step explanation:

1. The percent of voters who are other is the number of other divided by the total number of voters.

The number of "other" votes is 222. The number of total votes is 2,222.

The percent is 222/2222 = 0.0999. Times the decimal by 100 for the percent.

0.0999*100 = 9.99% rounds to 10%.

2. The probability is found by finding the number of male and registered as democrat, which is 600, and dividing it by the number of males, which is 1,222.

600/1,222 = 0.4909

Multiply by the decimal by 100 to find the percent.

0.4909*100 = 49.09 which rounds to 49.1%.

6 0

3 years ago

Point-slope of line's equation (-3,-7),(-8-11)

Nataly_w [17]

Steps:
1. Find the slope by using the slope formula
Y2-Y1
---------
X2-X1

-11 + 7
----------
-8 + 3
2.
-11 + 7= -4
-8 + 3= -5
3. Your slope is 4/5 (because you have to turn two negatives into a positive)
4. Now you do the point-slope formula
Y-Y1= m(x-x1)
Y +7=4/5 (x +3)

3 0

3 years ago

When attempting to do his homework, Mark incorrectly uses the addition property of equality below. Which

VikaD [51]

The correct answer is b

8 0

3 years ago

Your swimming pool containing 60,000 gal of water has been contaminated by 6 kg of a nontoxic dye that leaves a swimmer's skin a

Paul [167]

[a] Dye is removed from the pool at a rate of

(250 gal/min) * (q/60,000 g/gal) = -q/240 g/gal

where q denotes the amount of dye in the pool at time t. Clean water is pumped back into the pool, so no dye is being re-added.

So the net rate of change of the amount of dye in the pool is given by the differential equation,

$\dfrac{\mathrm dq}{\mathrm dt}=-\dfrac{q(t)}{240}$

with the intial value, q(0) = 6000 g (or 6 kg).

[b] The ODE above is separable as

$\dfrac{\mathrm dq}q=-\dfrac{\mathrm dt}{240}$

Integrate both sides to get

$\ln|q|=-\dfrac t{240}+C$

$e^{\ln|q|}=e^{-t/240+C}$

$\implies q(t)=e^{-t/240+C}=e^{-t/240}e^C=Ce^{-t/240}$

Now plug in the initial condition:

$6000=Ce^0\implies C=6000$

so the particular solution to the IVP is

$q(t)=6000e^{-t/240}$

[c] The acceptable concentration of the dy is 0.03 g/gal, which in a pool containing 60,000 gal of water corresponds to

(0.03 g/gal) * (60,000 gal) = 1800 g = 1.8 kg

of dye. Find the time t when this occurs:

$1800=6000e^{-t/240}\implies0.3=e^{-t/240}$

$\implies\ln0.3=-\dfrac t{240}$

$\implies t=-240\ln0.3\approx288.953$

so the amount of dye in the pool is within the acceptable tolerance after about 289 min have passes, or about 4.82 hrs. So no, the filtration system is not up to the task.

8 0

3 years ago

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