Solve 2cos^2x+3cosx-2=0

Answers

Answer 1
Answer: \bf 2cos^2(x)+3cos(x)-2=0\impliedby \textit{so, notice is just a quadratic} \n\n\n\ [2cos(x)~~-~~1][cos(x)~~+~~2]=0\n\n -------------------------------\n\n 2cos(x)-1=0\implies 2cos(x)=1\implies cos(x)=\cfrac{1}{2} \n\n\n \measuredangle x=cos^(-1)\left( (1)/(2) \right)\implies \measuredangle x= \begin{cases} (\pi )/(3)\n\n (5\pi )/(3) \end{cases}\n\n -------------------------------\n\n cos(x)+2=0\implies cos(x)=-2

now, for the second case, recall that the cosine is always a value between -1 and 1, so a -2 is just a way to say, such angle doesn't exist.

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What is bigger 1/5 or 2/10

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they are equal...1/5 is half of 2/10

What is the slope of side AC?

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Answer:

The slope is  ( d-b)/ ( c-a)

Step-by-step explanation:

To find the slope we can use the slope formula

m = ( y2-y1_/(x2-x1)

  = ( d-b)/ ( c-a)

Please
work out
∛15.76 - 4.6³ ÷2.5/3

Answers

Answer:

-10.47413972 or -10.47

Step-by-step explanation:

Calculator

Answer:

-10.47 is the answer

Step-by-step explanation:

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How do I put this in words? ​

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Answer: one hundred an twenty seven divided by uh hundred

Step-by-step explanation:

PLEASE HELP ME! THANK YOU!!

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It's obviously the second one

How many real-number solutions does the equation have? 1. –7x² + 6x + 3 = 0
a.one solution
b.two solutions
c.no solutions
d.infinitely many solutions

2.–8x² – 8x – 2 = 0
a.one solution
b.two solutions
c.no solutions
d. infinitely many solutions

Answers

If the discriminant is negative no real solution exists.
If the discriminant is equal to 0 only one real solution exists.
If the discriminant is positive 2 real solutions exist.
The discriminant:
D = b² - 4 a c
1 ) - 7 x² + 6 x + 3 = 0
D = 6² - 4 · ( - 7 ) · 3 = 36 + 84 = 120  > 0
Answer: b) two solutions
2 ) - 8 x² - 8 x - 2 = 0
D = ( - 8 )² - 4 · ( - 8 ) · ( - 2 ) = 64 - 64 = 0
Answer: a ) one solution