What is the square root of 1/36

Mathematics

1 answer:

lyudmila [28]3 years ago

6 0

Answer:

$\frac{1}{6}$

hope that helps uhh!!!

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Hey guys, I am about to start studying for a math test happening in two weeks about statistics and I am wondering if you guys ha

Stels [109]

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3 years ago

Use the diagram to find the measure of arc. AB and CD and the diameters of the circle F

dybincka [34]

Answer:

11). $m(\widehat{DE})$ = 90°

12). $m(\widehat{AEC})$ = 212°

Step-by-step explanation:

A circle F with AB and CD are the diameters has been given in the figure attached.

11). Since, $m(\widehat{AB})$ = 180°

and $m(\widehat{AB})=m(\widehat{AD})+m(\widehat{DE})+m(\widehat{BE})$

Therefore, $m(\widehat{AD})+m(\widehat{DE})+m(\widehat{BE})$ = 180°

32° + $m(\widehat{DE})$ + 58° = 180°

$m(\widehat{DE})$ = 180° - 90°

= 90°

12. Since, $m(\widehat{AD})=m(\widehat{BC})$ = 32°

$m(\widehat{AEC})$ = $m(\widehat{AB})+m(\widehat{BC})$

= 180° + 32°

= 212°

7 0

3 years ago

g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and

allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, $[ab]_K$ is equal to $[ba]_K$ , thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, $[ab]_H = [ba]_H$ . Since a and b were generic elements of H, then H/G is abelian.