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DedPeter [7]
3 years ago
10

Quick question if I had a problem that was sin x = 18.9/20 =.945 how would I inverse sin of both sides ?

Mathematics
1 answer:
galben [10]3 years ago
6 0

Answer:

The result is about 70.9 degrees

Step-by-step explanation:

Easy for a calculator.  If you're not using a calculator, it's much more tedious!

There is an inverse sine key on your calculator, but you may have to press a key (likely labeled 2nd or INV) first.  The attached image shows a TI-83 display.  To get the inverse sine, I first pressed the yellow 2nd key, then the sin key.

Check MODE first to make sure you're measuring angles in degrees (if that's what you want!).  Some calculators display angle mode with a small letter at the top of the display, D, R, or G usually.

The left side of the equation produces just x.

\sin{x}=.945\\\\\sin^{-1}(\sin{x})=\sin^{-1}(.945)\\x=\sin^{-1}(.945) \approx 70.9^\circ

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The roots of 2x^2-3x=4 are a and b. Find the simplest quadratic equation which has roots 1/a and 1/b
Tju [1.3M]
Find the roots

solve
we use hmm, completing the suare
2(x²-1.5x)=4
divide both sides by 2
x²-1.5x=2
take 1/2 of linear coeiftn and square it
-1.5/2=-0.75, (-0.75)²=0.5625
add that to both sides
x²-1.5x+0.5625=2+0.5625
factor perfect squaer trinomial
(x-0.75)²=2.5625
square root both sides, remember to take positive and negative square roots
x-0.75=+/-√2.5625
add 0.75 to both sides
x=0.75+/-√2.5625

the roots are x=0.75+√2.5625 and x=0.75-√2.5625

1/a and 1/b
1/(0.75+√2.5625) and 1/(0.75-√2.5625)

if the roots of a quadratic equation are r1 and r2 then it factors to
(x-r1)(x-r2)
so then we can factor our equation to be
(x-\frac{1}{0.75+\sqrt{2.5625}})(x-\frac{1}{0.75-\sqrt{2.5625}})

if we were to try and expand it, we would get
x²+0.75x-0.5
that's the simpliest equation with roots 1/a and 1/b where a and b are he roots of 2x²-3x=4


x²+0.75x-0.5 is answer
8 0
3 years ago
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