evaluate √16, if possible?

Question

also,
simplify 6√2 - 6√7
simplify √5 * √3
and
find the altitude of the triangle with sides L-13, R-13 and bottom 10.

Thanks for any help - I need it!

Answer 1

hmmm, as another person posted, your history of previous questions has a lot of math questions. Are you using this place to do your math homework?
Giving you answers won't help you learn. Giving you hints will help you figure it out for yourself.

To evaluate √16, your hint is:
Is there a number that when multiplied by itself equals 16? Think about it and you will be surprised at how simple this question is.

To simplify 6√2 - 6√7, your hint is:
To combine radicals, look for similarities either under the radical sign or in front of the radical sign. If the radicals are dissimilar but the numbers in front of the radicals are similar, then you can apply the Distributive Property.

To simplify √5 * √3, your hint is:
The rule for multiplying radicals is √a * √b = √(ab)

To find the altitude of the triangle, you have to split that triangle into two smaller triangles who share a side that is the unknown altitude value. Apply the pythagorean theorem to one of those triangles to calculate the unknown altitude side.

Good luck.
Have you considered asking your teacher for some tutoring?

Answer 2

√16 = +/- 4
simplify 6√2 - 6√7 = 6(√2 - √7)
simplify √5 * √3 = √15

find the altitude of the triangle with sides L-13, R-13 and bottom 10
=sqrt(13^-5^2) = 12

Answer 3

√16 = 4
Can't help with second one I'm afraid.
√5 * √3=√5*3=√15

For the triangle question use the cosine rule to find an angle.
CosA= (b^2+c^2-a^2)/(2*b*c). This gives you the bottom left angle as 67.38. Then use basic trig to work out the height.

The answer is 11.99998 or just 12

Answer 4

1st answer : 6(√2-√7)
2nd answer: try ur calculator... its 3.87298334520741688 & so on...
and 3rd answer: i didn't understand the question!!! sorry................

Answer 5

18
wait, this is your hmwk isn't it!i looked at your questio history and you keep asking these types of q's.