Show that the area of this right triangle is one half bc sine Upper A 1 2bcsinA. In right triangle A B C, the side opposite vertex A is labeled a, the side opposite vertex B is labeled b, and the side opposite vertex C is labeled c. Angle C is a right angle. The area of this right triangle is ▼ one half ac. 1 2ac. one half ab. 1 2ab. one half bc. 1 2bc. The base a of the right triangle ABC is ▼ c sine A. csinA. 2 c sine A.2csinA. 2 a cosine B.2acosB. a cosine B. acosB. Substituting aequals=c sine Upper AcsinA into the area formula gives the area of this right triangle is ▼ bc sine A. bcsinA. one half bc cosine B. 1 2bc cos B. one half bc sine A. 1 2bcsinA. bc cosine B.

the two may have that, even a breakdown is probably offered

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2