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sp2606 [1]
3 years ago
13

Determine the gcf. 2y^10, 5y^4, 6y^5

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

Answer:

y^4

Step-by-step explanation:

2y^10, 5y^4, 6y^5

We first need to find the factors of each term

2y^10 = 2*yyyyyyyyyy

5y^4 = 5yyyy

6y^5 = 3*2 yyyyy

The greatest common factor  is the largest term that appears in all the terms

The greatest common factor is yyyy  = y^4


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From his eye, which stands 1.59 meters above the ground, Francisco measures the
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The height of the skyscraper to the nearest tenth is 234. 95 meters.

<h3>How to determine the height</h3>

From the information given, we have:

  • angle of elevation = 32°
  • length of the base, adjacent side = 376 meters
  • height of the skyscraper, opposite side = x meters

To determine the height of the elevator, let's use the tangent identity

We have that;

tan θ = opposite/ adjacent

The value of the opposite side is 'x' which is the height of the skyscraper and the value of the adjacent side is 376 meters which is the length of the base of the skyscraper

Now, substitute the values, we have;

tan 32 = x/ 376

cross multiply

x = tan 32 × 376

x = 0. 6249 × 376

x = 234. 95 meters

We know that the 'x' represents the opposite side and thus the height of the skyscraper

Thus, the height of the skyscraper to the nearest tenth is 234. 95 meters.

Learn more about angle of elevation here:

brainly.com/question/19594654

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7 0
2 years ago
X+4 over 3 = x+6 over 4
Gnesinka [82]
If you are asking to solve for x the answer is:
x = 2
Hope This Helps!!
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4 0
2 years ago
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Suppose that a population parameter is 0.3 and many samples are taken from the population. If the size of each sample is 40, wha
scoundrel [369]

Answer:

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Step-by-step explanation:

A P E X

7 0
2 years ago
Find the expansion of cos x about the point x=0
ycow [4]

Answer:

Cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{1!} + ...

Step-by-step explanation:

We use Taylor series expansion to answer this question.

We have to find the expansion of cos x at x = 0

f(x) = cos x, f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, f''''(x) = cos x

Now we evaluate them at x = 0.

f(0) = 1, f'(0) = 0, f''(0) = -1, f'''(0) = 0, f''''(0) = 1

Now, by Taylor series expansion we have

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \frac{f''''(a)(x-a)^4}{4!} + ...

Putting a = 0 and all the values from above in the expansion, we get,

Cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{1!} + ...

8 0
3 years ago
Find the value of x. If necessary, round to the nearest tenth
andrew11 [14]
A = 1/2bh
40 = 1/2 x^2
x^2 = 80
    x = 8.9
3 0
3 years ago
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