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3 years ago

What does the |8| mean

Mathematics

2 answers:

Oxana [17]3 years ago

6 0

Answer:

ok so basically...idk how to explain it but i'll give u an example l-8l = 8 or l8l or l-6l = l6l, i don't know if u understand me, but basically it's like the.....opposite of that number or the absolute value

oksian1 [2.3K]3 years ago

5 0

Answer: The absolute value of 8

Step-by-step explanation:

Absolute value means how far away from 0 a number is.

If you have a number line, and the question is what is the absolute value of -10, for example, the answer would be ten, because it is 10 away from zero. if it was the absolute value of 10, however, it would still be ten, not -10.

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X Combine like terms: b + 4b2 - 8 + 3b - 2b2

emmainna [20.7K]

Answer:

$2b^{2}$ + 4b - 8

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3 years ago

What is the value of x? 3(x−2)+5x=18 X = ????

ikadub [295]

3(x-2)+5x=18

the correct answer is x=3

first simply the equation

8x=18+6

Now rid of alike terms resulting in

8x=24

now divide 24 by 8

x=3

6 0

3 years ago

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Choose 3 values for x that would make this inequality true. X - 4<20

Sergio039 [100]

Answer:

X= 23, 22, 21, 20, 19, 18, 17 and so forth

Step-by-step explanation:

To make this inequality true then "X-4" would have to be less than 20, not equal or more than. 24-4=20 meaning the basis for the answer of X would start at 23

6 0

3 years ago

Simplify: 8 x 7 + 4 17 + 8 x 2 20 x 3 + 10 5 + 4 x 15

belka [17]

Answer:

1.60

2.33

3.70

4.65

Step-by-step explanation:

Sana makatulong po ito!

7 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

Read 2 more answers