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Blizzard [7]
3 years ago
13

Find the least common multiple of these two expressions. 15x^7y^3u^8 and 6y^4u^5

Mathematics
1 answer:
snow_tiger [21]3 years ago
8 0

Answer:

30x^7y^4u^8

Step-by-step explanation:

We have been given two expressions 15x^7y^3u^8 and

6y^4u^5

Now we need to find out least common multiple of these two expressions.

First we need to find out what is common factor of both expressions. 15x^7y^3u^8 and

6y^4u^5

Least common multiple means find the expression which can be divided by both expressions.

15 and 6 both goes into 30

so 30 is part of LCM (least common multiple )

Now pickup the highest exponent of each variable.

So we get x^7y^4u^8

Hence required least common multiple is 30x^7y^4u^8

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When measuring forecast error, which forecasting error expresses the forecast error as a percentage from the mean?
oee [108]
Mean Absolute Percent Error, MAPE
4 0
3 years ago
What is 12/16 in simplest form
mojhsa [17]

Answer:

12/16 = 3/4

I divided both sides by 4.

Hope this helps!

-Mikayla


6 0
3 years ago
Read 2 more answers
Given f(x) = 6 (1-x) what is the value of f (-8)
Oxana [17]
Thanks for the question!

f(x) = 6 (1 - x)
f(-8) = 6 (1 - (-8))
f(-8) = 6 (9)
f(-8) = 54

Hope this helps!
3 0
3 years ago
What is the probability that the number of “HEADs” on these four coins is equal to 3? (4 points)
slavikrds [6]

Answer:

a. \frac{1}{4}  

Step-by-step explanation:  

We are asked to find the probability of getting 3 heads on 4 flips.

\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

Since we know that flipping a fair coin has 2 equally likely possible outcomes, so flipping four coins will have 2*2*2*2=16 possible outcomes.

Sample space of possible outcomes.

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,

THHH, THHT, THTH,THTT, TTHH, TTHT, TTTH, TTTT.

We can see that there are 4 favorable outcomes of getting heads.

\text{Probability of getting 3 heads}=\frac{4}{16}

\text{Probability of getting 3 heads}=\frac{1}{4}

Therefore, the probability of getting 3 heads on 4 coins will be \frac{1}{4} and option a is the correct choice.

5 0
3 years ago
Tickets to a football game cost $15.00 each including tax and a bag of popcorn cost $2.25 and a soda cost $1.75 Xavier has $350.
adoni [48]

Answer:

18 (full answer: 18.4210526316)

Step-by-step explanation:

<em>I LOST MY ANSWER RIGHT WHEN I WAS ABOUT TO FINISH DUE TO BRAINLY ROBOT DETECTION  </em>

This question seems incomplete. I dont know the tax, and what he wants to do with the $350, so I'll just work without them.

I'm going to assume that he wants to buy extra tickets, popcorn, and soda with the extra money.

1. add $15, $2.25, and $1.75

2. 15.00+2.25+1.75= 19.00

3. divide 350 by 19

350/19= roughly 18.42

So, Xavier can buy 18 tickets with food and drinks.

8 0
3 years ago
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