3 years ago

Px^2-qx+r=0 solve by complete squaring method

Mathematics

1 answer:

Natasha2012 [34]3 years ago

3 0

Step-by-step explanation:

px^2-qx+r

p(x^2-q/px)+r

p[(x^2-q/px +(q/2p)^2_(q/2p)^2)]+r

p[(x-q/2p)^2 -q^2/4p^2)]+r

p(x-q/2p)^2 -q^/4p+r

p(x-q/2p)^2-(q^2-4pr)/4p

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slamgirl [31]

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4 years ago

An angle in standard position measures radians, and P(0, 1) is on the terminal side of the angle. What is the value of the cosin

lana [24]

Answer: 0

Explanation:

1) Cosine trigonometric ratio definition:

cos (α) = adjacent-leg / hypotenuse

2) P (0,1) ⇒

- adyacent-leg = x-coordinate = 0
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3 years ago

Read 2 more answers

3(x + 3) + 10

Sphinxa [80]

Answer:

b-3x+19

Step-by-step explanation:

you have to open the bracket so 3*x=3x

3*3=9

therefore;3x+9+10=3x+19

7 0