What is the product? (6r-1)(- Br-3)

Mathematics

1 answer:

Andru [333]3 years ago

4 0

Answer:

$-6Br^2-18r+Br+3$

Step-by-step explanation:

we have the expression:

$(6r-1)(- Br-3)$

We need to multiply the two terms in the first parenthesis by the two terms in the second parenthesis:

$(6r)(-Br)+(6r)(-3)+(-1)(-Br)+(-1)(-3)$

we solve each term taking into account the law of signs

(-)(-)=+

and

(+)(-)=-

also for the first term we use that $r*r=r^2$

thus we have:

$-6Br^2-18r+Br+3$

it can be simplified a little more if we factor an r of the second and third term:

$-6Br^2+r(-18+B)+3$

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If a = 7 and b = 11, what is the measure of ∠B? (round to the nearest tenth of a degree) A) 32.5° B) 39.2° C) 50.5° D) 57.5°

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D) 57.5°

Step-by-step explanation:

As the question is not complete. So, let's suppose it is a right angle triangle then, we can apply Pythagoras theorem to calculate the hypotenuse or the third side.

Pythagoras Theorem = $c^{2}$ = $a^{2} + b^{2}$