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Ivahew [28]
2 years ago
5

What is the distance formula?

Mathematics
1 answer:
marin [14]2 years ago
8 0
Your answer is D.
Will always be x2 and y2 starting off
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Which fraction is easier to convert into a decimal?<br><br> 11/18 or 43/50, explain
alexgriva [62]

43/50 is easier because 50 is half of 100

Watch this

43/50. Since 50 is HALF of 100 so we multiply 2 by 50 which is 100. But since we multiplied by 2 for the denominator we have to do the same for the numerator so 43 x 2 is equal to 86.

So we get 86/100 which is 0.86

5 0
2 years ago
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You are renting a house in Portland for a week at $1000. What is the cost per day? Tell me your exact steps.
AlexFokin [52]

Answer:

142.85

Step-by-step explanation:

1000 divided by 7

6 0
3 years ago
Task: Nonpermissible Values for Rational Expressions [30 points] Write a rational expression that has the nonpermissible values
igor_vitrenko [27]

Answer:

A rational expression that has the nonpermissible values x = 0 and x = 17 is f(x) = \frac{4}{x\cdot (x-17)}.

Step-by-step explanation:

A rational expression has a nonpermissible value when for a given value of x, the denominator is equal to zero. In addition, we assume that both numerator and denominator are represented by polynomials, such that:

f(x) = \frac{p(x)}{q(x)} (1)

Then, the factorized form of q(x) must be:

q(x) = x\cdot (x-17) (2)

If we know that p(x) = 4, then the rational expression is:

f(x) = \frac{4}{x\cdot (x-17)} (3)

A rational expression that has the nonpermissible values x = 0 and x = 17 is f(x) = \frac{4}{x\cdot (x-17)}.

8 0
3 years ago
Tom needs to drive 21 miles to work. So far, he has driven 2.2 miles. How many more miles must he drive?
wolverine [178]
Tom needs to drive 18.8 more miles
4 0
3 years ago
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g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and
allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, [ab]_K is equal to [ba]_K, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, [ab]_H = [ba]_H . Since a and b were generic elements of H, then H/G is abelian.

4 0
3 years ago
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