Find the GCF of 6fgh^2, 3f^2h, 12g^2h.

Mathematics

1 answer:

Mariulka [41]3 years ago

8 0

The gcf are (8) (2) and (18)

You might be interested in

Find the value of each variable

aniked [119]

Answer:

Answer d)

$a= 10*\sqrt{3}$ $b=5*\sqrt{3}$ , $c=15$ , and $d=5$

Step-by-step explanation:

Notice that there are basically two right angle triangles to examine: a smaller one in size on the right and a larger one on the left, and both share side "b".

So we proceed to find the value of "b" by noticing that it the side "opposite side to angle 60 degrees" in the triangle of the right (the one with hypotenuse = 10). So we can use the sine function to find its value:

$b=10*sin(60^o)= 10*\frac{\sqrt{3} }{2} = 5*\sqrt{3}$

where we use the fact that the sine of 60 degrees can be written as: $\frac{\sqrt{3} }{2}$

We can also find the value of "d" in that same small triangle, using the cosine function of 60 degrees:

$d=10*cos(60^o)=10* \frac{1}{2} = 5$

In order to find the value of side "a", we use the right angle triangle on the left, noticing that "a" s the hypotenuse of that triangle, and our (now known) side "b" is the opposite to the 30 degree angle. We use here the definition of sine of an angle as the quotient between the opposite side and the hypotenuse:

$sin(30^o)= \frac{b}{a} \\a=\frac{b}{sin(30)} \\a=\frac{5*\sqrt{3} }{\frac{1}{2} } \\a= 10*\sqrt{3}$

where we used the value of the sine function of 30 degrees as one half: $\frac{1}{2}$

Finally, we can find the value of the fourth unknown: "c", by using the cos of 30 degrees and the now known value of the hypotenuse in that left triangle:

$c=10*\sqrt{3} * cos(30^o)=10*\sqrt{3} *\frac{\sqrt{3} }{2} \\c= 5*3=15$