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3 years ago

6

How can i recognize a Absolute value?

2 answers:

Lunna [17]3 years ago

6 0

First of all you have to count how far away that number is from 0 (Zero) on a number line, and thats your answer, but be sure to remember, Absolute Value can never be negative!

postnew [5]3 years ago

4 0

You can if the number has l l next to it

for example: l 2 l

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A pitcher of water contains 1/4 of water. The water is poured equally into 5 glasses. How many liters of water are in each glass

IRINA_888 [86]

Hey there!

To solve this, divide the numbers 1/4 and 5 . . .

1/4 = 1/4

5 = 5/1

1 x 1 = 1

4 x 5 = 20

= 1/20

1/20 is already in it's simplest form, so it doesn't need to be simplified.

There are 1/20 liters of water in each glass.

Hope this helps you.
Have a great day!

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3 years ago

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I really need help on this question!!!

Marysya12 [62]

Answer: 1795.

Step-by-step explanation:

The statement states that it was invented "177 years before".

It also states the year the video game was invented which is 1972.

1972-177=1795.

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3 years ago

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Calculate the slope, or rate of change, of this line.<br><br> 1) 3<br> 2) 4<br> 3) 2<br> 4) 0.5

ASHA 777 [7]

Answer:

The slope for the line is 2

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3 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

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2 years ago

I not sure what it is? Thank you to who ever help me

Vadim26 [7]

I think b I might be wrong

8 0

3 years ago