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Molodets [167]
4 years ago
8

Write the equation of the line that passes through (1, 3) and has a slope of 2 in point-slope form. A) y – 1 = 2(x – 3) B) y – 3

= 2(x – 1) C) x – 1 = 2(y – 3) D) x – 3 = 2(y – 1)
Mathematics
1 answer:
fomenos4 years ago
5 0
So point slope form is y=mx+b where m=slope and b=y intercept so

in (x,y) form
one solution is x=1 and y=3, also slope is 2 so
m=2
subsitute
3=2(1)+b
3=2+b
subtract 2
1=b
equation is
y=2x+1

tricky answers so simplify them
A: y-1=2(x-3)
distribute
y-1=2x-6
add 1
y=2x-5
not the answer

B: y-3=2(x-1)
distribute
y-3=2x-2
add 3
y=2x+1
correct
the answer is B
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Multiply the polynomials.
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Answer:

The answer to your question is: letter A

Step-by-step explanation:

      (x + 3) (3x² + 8x + 9) = 3x³  + 8x² + 9x + 9x² + 24x + 27

                                       = 3x³ + 17x² + 33x + 27              Simplify like terms      

Α. 3x3 + 17x2 + 33x + 27

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​

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3 years ago
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Read 2 more answers
What is the exact value of cos (67.5°)?
babymother [125]

first off, make sure you have a Unit Circle, if you don't do get one, you'll need it, you can find many online.

let's double up 67.5°, that way we can use the half-angle identity for the cosine of it, so hmmm twice 67.5 is simply 135°, keeping in mind that 135° is really 90° + 45°, and that whilst 135° is on the 2nd Quadrant and its cosine is negative 67.5° is on the 1st Quadrant where cosine is positive, so

cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\[-0.35em] ~\dotfill\\\\ cos(135^o)\implies cos(90^o+45^o)\implies cos(90^o)cos(45^o)~~ - ~~sin(90^o)sin(45^o) \\\\\\ \left( 0 \right)\left( \cfrac{\sqrt{2}}{2} \right)~~ - ~~\left( 1\right)\left( \cfrac{\sqrt{2}}{2} \right)\implies -\cfrac{\sqrt{2}}{2} \\\\[-0.35em] ~\dotfill

cos(67.5^o)\implies cos\left( \frac{135^o}{2} \right)\implies \pm \sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}\implies \stackrel{I~Quadrant}{+\sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}} \\\\\\ \sqrt{\cfrac{ ~~ \frac{2-\sqrt{2}}{2} ~~ }{2}}\implies \sqrt{\cfrac{2-\sqrt{2}}{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{2}

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