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Rashid [163]
2 years ago
9

Please help if you can! :)

Mathematics
1 answer:
mina [271]2 years ago
5 0

9514 1404 393

Answer:

  A. no solution

  B. unique solution: (x, y) = (0, 0)

Step-by-step explanation:

For these problems, you can subtract the second equation from the first to get something of the form ...

  0 = (expression)

The solution of that will tell you the answer.

__

A. (2x +3y) -(2x +3y) = (4) -(9)

  0 = -5 . . . . . . . . no values of x or y will make this true: NO SOLUTION

__

B. (y) -(y) = (6x) -(x)

  0 = 5x

  0 = x . . . . . . divide by the coefficient of x.

  y = x = 0

The unique solution is (x, y) = (0, 0).

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Answer lol i dont know

Step-by-step explanation:

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Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1
IceJOKER [234]

Solution to equationcosxtanx - 2 cos^2 x=-1 for all real values of x is  x=2k\pi + \frac{\pi}{6}  , x=2k\pi + \frac{5\pi}{6} .

<u>Step-by-step explanation:</u>

Here we have , cosxtanx - 2 cos^2 x=-1. Let's solve :

⇒  cosxtanx - 2 cos^2 x=-1

⇒  cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1

⇒  sinx = 2 cos^2 x-1

⇒  sinx = 2 (1-sin^2x)-1

⇒  sinx = 1-2sin^2x

⇒  2sin^2x+sinx-1=0

By quadratic formula :

⇒ sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}

⇒ sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}

⇒ sinx = \frac{-1 \pm3}{4}

⇒ sinx = \frac{1}{2} , sinx =-1

⇒ sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}

⇒ x=\frac{\pi}{6} , x=\frac{3\pi}{2}

But at x=\frac{3\pi}{2} we have equation undefined as cos\frac{3\pi}{2}=0 . Hence only solution is :

⇒ x=\frac{\pi}{6}

Since , sin(\pi -x)=sinx

⇒ x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}

Now , General Solution is given by :

⇒ x=2k\pi + \frac{\pi}{6}  , x=2k\pi + \frac{5\pi}{6}

Therefore , Solution to equationcosxtanx - 2 cos^2 x=-1 for all real values of x is  x=2k\pi + \frac{\pi}{6}  , x=2k\pi + \frac{5\pi}{6} .

3 0
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The answer is d and because if you divide 15 by 5 it’s 3 so it’s 9
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