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DiKsa [7]
1 year ago
14

Explain how to identify the restrictions on the variable when adding or

Mathematics
1 answer:
anastassius [24]1 year ago
3 0

Rational expressions are those that have fractional terms. We state restrictions because it may cause the equation to be undefined in some values of x. The most common restriction for rational expressions is N/0. This means any number divided by zero is undefined.

<h3>What is rational expressions and examples?</h3>

A rational expression is nothing more than a fraction in which the numerator and/or the denominator are polynomials. Here are some examples of rational expressions.

To find the restrictions on a rational function, find the values of the variable that make the denominator equal 0.

  • 6x-1z2-1z2+5m4+18m+1m2-m-64*2+6*-101

Learn more about rational expressions here:

brainly.com/question/9353162

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Please I need help. you will get points ​
iragen [17]

Answer:

t(32) = 94

Step-by-step explanation:

a. I think s is the independent variable

b I think t or t(s) is the dependent variable since the total amount of money depends on how many treats are sold.

t(32) = 30 + 2(32)

t(32) = 30 + 64

t(32) = 94

I'm a little unsure on the dependent and independent variable but I know the function is correct.

Hope this helps :)

7 0
3 years ago
Margaret bought 1,575 grams of rice. how many kg of rice is that?
bearhunter [10]
1 kg = 1000 grams

? kg = 1575 grams

1.575 kg = 1575 grams

1,575 grams is 1.575 kilograms.
6 0
3 years ago
35 of of 30 in simplest form
Mariana [72]

Answer:7/6

Step-by-step explanation:

divide both 35 & 30 by 5 since it’s their greatest common factor.

that would get u 7/6

8 0
2 years ago
NEED HELP ASAP.<br> Check my answer
lisabon 2012 [21]

that looks like a regular octagon to me

6 0
3 years ago
Read 2 more answers
Help asap please!!!!
Lemur [1.5K]

Answer:

( -7.5, 3.5)

Step-by-step explanation:

M=(\frac{x1+x2}{2} ,\frac{y1+y2}{2} )\\\\M=(\frac{-5-10}{2} ,\frac{5+2}{2} )\\\\M=(\frac{-15}{2} ,\frac{7}{2} )\\\\M=(-7.5,3.5)

7 0
3 years ago
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