3 years ago

Help with khan academy problem, law of sines

1 answer:

7 0

Answer:

104°

Step-by-step explanation:

$\frac{sin~B}{18} =\frac{sin~29}{9} \\sin~B=\frac{18}{9} \times sin~29\\sin~B=2~sin~29\\B=sin^{-1}(2 ~sin~29)\\=104^\circ$

You might be interested in

VMariaS [17]

It would be a a 6 by 5 inch rectangle, since the height is 6in and the width is 5in. Those are what would make up the cross section.

6 0

3 years ago

antiseptic1488 [7]

Answer:

A... .................. ....

4 0

3 years ago

allochka39001 [22]

I hope this helps you

3 0

3 years ago

Shtirlitz [24]

Answer:

It should be B $250 - - - - -

3 0

3 years ago

blsea [12.9K]

Answer:

Step-by-step explanation:

$y^2-11y+n\\=(y-(11/2))^2+n-(11/2)^2$

So in order the trinomial a perfect square we need

$n-(11/2)^2 =0\\n= (11/2)^2\\n=\frac{121}{4}$

When n is equal to this number that we have

$y^2-11y+n = (y- (11/2))^2$

6 0

2 years ago