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andreyandreev [35.5K]
3 years ago
9

What is the gcf of 99 and 54

Mathematics
2 answers:
Sergio [31]3 years ago
6 0

Answer:

9

Step-by-step explanation:

The factors of 99 are 99, 33, 11, 9, 3, 1. The common factors of 54 and 99 are 9, 3, 1, intersecting the two sets above. In the intersection factors of 54 ∩ factors of 99 the greatest element is 9. Therefore, the greatest common factor of 54 and 99 is 9.

Hope this helps :)

Can I have brainliest?

maksim [4K]3 years ago
5 0

Answer:

9

Step-by-step explanation:

9 x 11 = 99

9 x 6 = 54

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If it takes 5 1/7 yards of material to make a pillow cover, how much material will be needed to make 9 pillow covers?
s2008m [1.1K]

\text{Answer: }46\frac{2}{7}\text{ yards is needed to make 9 pillow covers.}

Step-by-step explanation:

Since we have given that

Material needed to make a pillow cover is given by

5\frac{1}{7}\ yards=\frac{36}{7}\ yards

We need to calculate the material which will be needed to make 9 pillows covers.

So, we use the "Multiplication Operation" as we need the value for more pillow covers :

9\times \frac{36}{7}\\\\=\frac{324}{7}\\\\=46\frac{2}{7}\ yards

\text{Hence, }46\frac{2}{7}\text{ yards is needed to make 9 pillow covers.}

3 0
3 years ago
Below are several points on the graph of a polynomial. Which of the following statements are true about the graph of this polyno
FinnZ [79.3K]

Notice that we need to look for the intervals of x and also for the intervals of y.<span> We can observe that t</span>here is a minimum between x = 0 and x = 1 because the graph is descending and then ascending and also the graph has a maximum value between y = -3 and y = -1.

4 0
3 years ago
Fission tracks are trails found in uranium-bearing minerals, left by fragments released during fission events. An article report
Harlamova29_29 [7]

Answer:

Mean track length for this rock specimen is between 10.463 and 13.537

Step-by-step explanation:

99% confidence interval for the mean track length for rock specimen can be calculated using the formula:

M±\frac{t*s}{\sqrt{N}} where

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  • t is the two tailed t-score in 99% confidence interval (2.977)
  • s is the standard deviation of track lengths in the report (2 μm)
  • N is the total number of tracks (15)

putting these numbers in the formula, we get confidence interval in 99% confidence as:

12±\frac{2.977*2}{\sqrt{15}} =12±1.537

Therefore, mean track length for this rock specimen is between 10.463 and 13.537

4 0
3 years ago
Please help fast ^^
allsm [11]
The discriminante :
b^2-4ac

1^2 - 4 * -2 * -28 = 1 - 224 = -223

When the discriminant (b^2-4ac) is less than 0, the equation had no real solutions.

-223<0, so, 2x^2+x-28 = 0 has no real solutions.


Hope that helps :)
7 0
2 years ago
Mary is traveling cross country on a road trip. So far, Mary has traveled 200 miles in her car and 200 miles by train. The car t
VikaD [51]

Answer:

i dont know i think it is 11?!? srry if im wrong

Step-by-step explanation:

5 0
3 years ago
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