Find the terminal point on the unit circle determined by

Mathematics

1 answer:

aalyn [17]3 years ago

4 0

Use Conects it’s a better app for math hope this helps

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Rewrite the expression $6j^2 - 4j 12$ in the form $c(j p)^2 q$, where $c$, $p$, and $q$ are constants. What is $\frac{q}{p}$

solniwko [45]

$Rewrite the expression 6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, where $c$, $p$, and $q$ are constants. What is $\frac{q}{p}$$

The ratio of $\frac{q}{p}$ = - 34

How to solve such questions?

Such Questions can be easily solved just by some Algebraic manipulations and simplifications. We just try to make our expression in the form which question asks us. This is the best method to solve such questions as it will definitely lead us to correct answers. One such method is completing the square method.

Completing the square is a method that is used for converting a quadratic expression of the form $ax^{2} + bx + c$ to the vertex form

$a(x - h)^{2} + k$ . The most common application of completing the square is in solving a quadratic equation. This can be done by rearranging the expression obtained after completing the square: $a(x + m)^{2} + n$ , such that the left side is a perfect square trinomial

$6j^2 - 4j +12$

= $6(j^2 - \frac{2}{3} j )+12$

= $$6(j^2 - \frac{2}{3} j + \frac{1}{9} )+\frac{102}{9}$ (Completing Square method)

= $6( j- \frac{1}{3} )^{2} + \frac{34}{3}$

On comparing with the given equation we get

p = - $\frac{1}{3}$ and q = $\frac{34}{3}$

∴ $\frac{q}{p}$ = $\frac{\frac{34}{3} }{\frac{-1}{3} }$

= - 34

Learn more about completing the square method here :

brainly.com/question/26107616

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2 years ago

Need help on this question

stellarik [79]

Answer:nhjhbghhnkjn hojn

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3 years ago

Read 2 more answers

1<br>Prove that<br>1: sin/1-cot + cos/1-tan=cos+sin

Anit [1.1K]

Answer:

have:

$\frac{sin}{1-cot}+\frac{cos}{1-tan}\\\\=\frac{sin}{1-\frac{cos}{sin} }+\frac{cos}{1-\frac{sin}{cos} }\\\\=\frac{sin^{2} }{sin-cos}+\frac{cos^{2} }{cos-sin} \\\\=\frac{sin^{2} }{sin-cos}-\frac{cos^{2} }{cos-sin}\\\\=\frac{sin^{2}-cos^{2} }{sin - cos}\\\\=\frac{(sin-cos)(sin+cos)}{sin-cos}\\\\=sin+cos$