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bearhunter [10]
1 year ago
13

What is the GCF of 16s3t, 40s5, and 68t2? 4 4s3t 8 8s3t

Mathematics
2 answers:
kumpel [21]1 year ago
9 0

Answer:  First option is correct.

Step-by-step explanation:

Since we have given that

16s^3t,40s^5\ and\ 68t^2

First we write the factors of all of these numbers:

Factors of 16s^3t is given by

4\times 4\times s\times s\times s\times t

Factors of 40s^5 is given by

2\times 2\times 2\times 5\times s\times s\times s\times s\times s

Factors of 68t^2 is given by

17\times 2\times 2\times t\times t

So, GCF ( Greatest common factor) will be

2\times 2=4

Hence, First option is correct.

Gemiola [76]1 year ago
3 0
4 is the only factor common among all terms
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Right angle FCD intersects Line A B and Ray C E at point C. AngleFCE is congruent to AngleECD. AngleECD is complementary to Angl
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Answer:

A: They are congruent and complementary.

Step-by-step explanation:

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1 year ago
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​mars, inc. claims that​ 20% of its​ m&m plain candies are orange. a sample of 100 such candies is randomly selected. find t
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The probability p of an orangecandy is 0.2. The sample size = 100.
The mean is given by:
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The standard deviation is given by:
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2 years ago
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Find any relative extrema of the function. (Round your answers to three decimal places.)
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(x) = arcsec(x) − 8x

f'(x) = d/dx( arcsec(x) − 8x )

<span> 1/xsqrt( x^2 - 1) - 8</span>

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1/xsqrt( x^2 - 1) - 8 = 0

8 x sqrt (x^2-1) = 1

<span> ( 8 x sqrt (x^2-1) )^2 = 1</span>

64 x^2 ( x^2 - 1) = 1

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x = 1.00766 , - 1.00766

<span> x =   - 1.00766</span>

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8 0
2 years ago
James is the manager at an entertainment arena that draws an average 7,000 patrons per event. Each ticket taker can process 350
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Answer:

No. James didn't have enough ticket takers to process the average number of patrons hat usually attend the events.

He need to hire 2 more ticket taker (i.e 20 ticket takers) in order to process the average number of patrons hat usually attend the events.

Step-by-step explanation:

Given:

Average amount of patron per event= 7000

Each ticket taker can process = 350

Number of ticket takers hired = 18

We need to find the whether he hire enough to process the average number of patrons that usually attend the events.

Solution:

We will find the Average amount of patron ticket takers can process.

Now we can say that

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framing in equation form we get;

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Hence With the required hires James cannot fully process the patrons usually attending the event.

To find how many ticket takers required we will divide average number of patrons with  Each ticket taker can process.

framing in equation form we get;

ticket takers required = \frac{7000}{350}=20

hence In order t process the the average number of patrons that usually attend the events James will require 20 ticket takers.

6 0
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=100.2743338823 which is 100
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