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Sholpan [36]
2 years ago
13

How do we solve the three variable systems

Mathematics
1 answer:
zaharov [31]2 years ago
3 0

Answer:

Step-by-step explanation:

1. Pick any two pairs of equations from the system.

2. Eliminate the same variable from each pair using the Addition/Subtraction method.

3. Solve the system of the two new equations using the Addition/Subtraction method.

<u>Please mark Branliest!</u>

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Put these in order from greatest to least: 2/5 5/6 5/8 1/2 3/7
Leya [2.2K]

Answer:

2/5, 3/7,  1/2, 5/8, 5/6

Step-by-step explanation:

change the fractions into decimals and place them in order from least to greatest, then revert back to it's original form which would be a fraction.

4 0
3 years ago
Inverse function for y=x^2
Leviafan [203]

Answer:

y= \sqrt{x}

Step-by-step explanation:

To find inverse of something, just switch the variables

x=y^2

now try to solve back in terms of y

y= \sqrt{x}

3 0
4 years ago
Find the distance traveled by driving at 55 mph for 3 hours.
Zolol [24]
You just need to divide

5 0
3 years ago
Read 2 more answers
Jane wrote the numbers 1-100 and placed them in a bag. If she selects a number without looking, what is the probability that the
Sloan [31]

Since she wrote the numbers 1-100, we know that the probability for one number is 1/100.

If she selects a number without looking, it's a 1/100 chance she'll pull out one specific number.

However, here we need to find the probability of a number below 5. There are four numbers below 5 from the range to 1-100 and they are 1, 2, 3, and 4.

Therefore, since there are 4 numbers below 5, we have a 4/100 chance of Jane pulling out a number below 5.

In simplified terms it would be \frac{1}{25}

In decimal form, it would be 0.04

7 0
3 years ago
Limit xtens to 0 x^2logx^2 what is the ans of interminate forms?
GarryVolchara [31]

Rewrite the limit as

\displaystyle\lim_{x\to0}x^2\log x^2=\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}

Then both numerator and denominator approach infinity (with different signs, but that's not important). Applying L'Hopital's rule, we get

\displaystyle\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}=\lim_{x\to0}\frac{\frac2x}{-\frac2{x^3}}=\lim_{x\to0}-x^2=\boxed{0}

7 0
3 years ago
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