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e-lub [12.9K]
3 years ago
9

Math help, BOTH of these please!

Mathematics
2 answers:
Feliz [49]3 years ago
8 0
#1 : ( 3 , - 2 )

~~

#2 : ( - 1 , 4 )

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
kirill115 [55]3 years ago
4 0
1. You are correct
2. (4, - 5)
Hope this helps you if you want me to explain why I will be happy to!
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D. 46.5%

Step-by-step explanation:

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Help help<br><br> Order these fractions from least to greatest.
Dmitrij [34]

Answer:

\frac{5}{9}, \frac{7}{9}, \frac{8}{9}

Step-by-step explanation:

The answer is: \frac{5}{9}, \frac{7}{9}, \frac{8}{9}

And this question isn't to difficult, and the trick is realizing they all have the same denominator.

The denominator is essentially saying how many pieces of the whole (1) is being broken into.

So the denominator 9, means that you're breaking up the number 1 into 9 pieces.

The numerator is telling us how many pieces of this "broken up 1" the number has.

So 8/9 has more of these "broken up pieces" than 7/9, and it's so vitally important that these numbers have the same denominator.

For example: 1/2 and 30/90

30/90 may appear to be bigger, since the numerator of 30, but it's actually smaller.

The reason for this is because we can simplify it to: 1/3 by dividing both sides by 30 so the value remains the same.

Now we can actually compare 1/2 and 1/3 without changing the denominators.

1/3 is "one piece" of a broken up one, that was "broken" into 3 pieces.

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8 0
1 year ago
How many ways are there to select five unordered elements from a set with three elements when repetition is allowed?
Zolol [24]

The number of ways or permutations to select five unordered elements from a set with three elements when repetition is allowed is 243.

According to the given question.

The total number of elements, n = 3

The number of elements to be selected at a time when repetition is  allowed, r = 5

Since, we know that " the number of ways or permutations of n things taken r all at a time, when repetition of things are allowed, is n^{r}.

Therefore,

The number of ways or permutations to select five unordered elements from a set with three elements when repetition is allowed

= 3^{5}

= 243

Hence, the number of ways or permutations to select five unordered elements from a set with three elements when repetition is allowed is 243.

Find out more information about permuations here:

brainly.com/question/13833229

#SPJ4

3 0
1 year ago
A pound of ground beef costs $5. At this rate, what is the cost of:
jeka94
Yes 12-14 is the answer hope this helps !!!
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