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nikklg [1K]
4 years ago
7

A proportional relationship can be shown in two different ways:​

Mathematics
1 answer:
Vladimir [108]4 years ago
6 0
What are u asking ? I’m don’t get it
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I HAVE 5 MIN LEFT PLEASE HELP
Murljashka [212]

13.5 * 12.2 = 164.7 i think that's the answer

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3 years ago
Which fraction is equivalent to <br><br> 3/5
pentagon [3]

Answer:

12/20

Step-by-step explanation:

5 0
3 years ago
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Geometry: complete this proof of theorem 14, ASAP!
PtichkaEL [24]

Answer:

1. Transversal y intersects lines m and n; <1 ~= <2 (Given)

2. <1 ~= <3 (Vertical Angles Theorem)

3. <2 ~= <3 (Transitive Property of Congruence)

4. m || n (Converse of Alternate Interior Angles Theorem)

3 0
2 years ago
Please help me with the article help please help me
Anvisha [2.4K]

Answer:

I think it's like this

Article -6

Description: coffee........ gunguli .

who is interested in statistics?

-karen riche at the French National institute of medical research ,whose work appears in journal neurology

who is the audience?

-7000 men's and women's in the three french cities.

what kind of decision one should make after reading this article?

reading the above article I came to acknowledge that women are only benefited as they don't metabolizes coffee in same way as men.so men should take more care and control over their coffee consumption than women.

Description: coffee........ gunguli

(in here write whole sentence I just did in short cut)

8 0
3 years ago
Give an example of a function f : N → N that is surjective but not injective. You must explain why your example is surjective an
Helen [10]

Answer:

Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.

Step-by-step explanation:

First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y.  We conclude that f is surjective.

However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.

Note:

If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.

7 0
3 years ago
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