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3 years ago

The GCF of 28 and 64 is 2 4 7 8

2 answers:

8 0

The greatest comment factor of 28 and 64 is 4.

The factors of 28 are 28, 14, 7, 4, 2, 1.
The factors of 64 are 64, 32, 16, 8, 4, 2, 1.

The common factors of 28 and 64 are 4, 2.

the greatest one of the two factors is 4.

ad-work [718]3 years ago

3 0

Another way to find GCF is by the euclidean algorithm. Even though the name sounds intimidating, it is probably the easiest method, especially if the numbers are large.

The method consists of calculating the difference of the two given numbers. Then among the two numbers and the difference, repeat with the two smaller numbers until the difference is zero. The smallest surviving number is then the GCF.

As an example for 64 and 28.

64-28=36. Choose the smaller two numbers and repeat

36-28=8 Choose the smaller two numbers and repeat

28-8=20 Choose the smaller two numbers and repeat

20-8=12 Choose the smaller two numbers and repeat

12-8=4 Choose the smaller two numbers and repeat

8-2*4=0 therefore 4 is the GCF.

As a short-cut, when the larger number is much larger than the smaller, we can subtract a multiple of the smaller number. Example again using 64 and 28.

64-28=36

36-28=8

28-3*8=28-24=4

8-2*4=0 Therefore 4 is the GCF of 64 and 28.

As a more adventurous example, let's find the GCF of 3888 and 3774.

3888-3774=114

3774-30*114=354

354-3*114=12

12-8*12=6

12-2*6=0 Therefore 6 is the GCF of 3888 and 3774.

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There are two halves in an hour so

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4 years ago

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I need help with finding the answer to a) and b). Thank you!

shtirl [24]

Answer:

$\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }$

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}$

$\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin(x)=-\frac{21}{29}$

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

$a=\sqrt{29^2-21^2}=20$

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

$180$

So:

$\displaystyle 90 < \frac{x}{2} < 135$

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

Recall that:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}$

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 + 20/29}{2}$

Simplify:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}$

Likewise:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }$

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }$

Simplify:

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}$

Finally:

$\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}$

Therefore:

$\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}$

5 0

3 years ago