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ruslelena [56]
3 years ago
12

When lava is expelled from a volcano, it temperature may vary from 700 degrees Celsius to 1200 degrees Celsius. Write an absolut

e value equation that represents the minimum and maximum temperatures of lava. Use x to represent the temperature.
Equation:__
Mathematics
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

Ix - 950°C I ≤ 250°C

Step-by-step explanation:

We are told that the temperature may vary from 700 degrees Celsius to 1200 degrees Celsius.

And that this temperature is x.

This means that the minimum value of x is 700°C while maximum of x is 1200 °C

Let's find the average of the two temperature limits given:

x_avg = (700 + 1200)/2 =

x_avg = 1900/2

x_avg = 950 °C

Now let's find the distance between the average and either maximum or minimum.

d_avg = (1200 - 700)/2

d_avg = 500/2

d_avg = 250°C.

Now absolute value equation will be in the form of;

Ix - x_avgI ≤ d_avg

Thus;

Ix - 950°C I ≤ 250°C

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Hese are the values in Paul’s data set.
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sveticcg [70]

Answer to 1:

\frac{x^3}{2y^25}

First thing you want to do is combine like terms on the bottom. Since it's multiplication, all you have to do is add the exponents. So for x you'd get x raised to the -9 power and for y you'd get 2y raised to the power of 7

Second thing is the numerator of the fraction. Since this is raising what is inside the paranthesis to a power, you must multiply this power by the powers of x and y, respectively. You should get x raised to the power of -6 and y raised to the power of -18

Here's where it gets complicated. to rationalize the problem, you must add parts from the numerator and denominator or vice versa. You cannot have any negative exponents anywhere in the fraction.  So that means for x you have to add 9 to the x on the top to cancel out the -6 and rationalize the remaining x's. You should end up with x^3.

For y, you work the opposite way. Since the negatives are only on the top, you simply add 18 to the bottom y's and it remains rationalized.

If done correctly this should come out to x^3/2y^25

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(x^20y^12)/256

First thing you can do on this one is to cancel out the x^0s. Anything to the power of 0 equals one so only the coefficients will remain. Multiply those together and you'll get 4.

Next you will rationalize the ys. Add the bottom 3 to the top to cancel out the denominator and you're left with y^-3. Leave it like this for now.

Now is the difficult part. You must take everything that is inside the parenthesis to the power of -4. That includes the coefficients and the exponents. 4^-4 = 1/256. This means that you now must move the coefficients to the bottom. You should currently have y^-3/(256x^5).

Now take the exponents to the power using the same rules as question 1. -3*-4 = 12 and 5*-4 = -20. You should now have y^12/(256x^-20)

Lastly, we must rationalize the denominator. To do so, move the x^-20 to the numerator and make the exponent positive. After doing this, you have the answer: (x^20y^12)/256

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