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s344n2d4d5 [400]
3 years ago
11

Hi can someone please help me with these two questions ive been struggling solving and explaining how i found the answer and fin

ding the answer for them.
1) 28-6x+4=30-3x

2)4x-1=9x-1-5x​
Mathematics
1 answer:
DIA [1.3K]3 years ago
3 0

Answer:

x = 2/3; all real numbers.

Step-by-step explanation:

                       • hey! •

let's move step-by-step to evaluate these equations.

1.  28 - 6x + 4 = 30 - 3x

- group like terms -

-6x+28+4=30-3x

- add! -

-6x+32=30-3x

- subtracting and simplifying -

-6x=-3x-2

- simplify -

\frac{-3x}{-3}=\frac{-2}{-3}

\boxed{x = \frac{2}{3} }.

2. 4x - 1 = 9x - 1 - 5x​

- add -

4x-1+1=9x-1-5x+1

- simplify -

4x=9x-5x

- add -

4x=4x

thusly, \boxed{{\text{all real numbers}}} are solutions to the equation. :)

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sergeinik [125]

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So from definition of tan we have this:

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a. sin (17pi / 4 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to 2\pi:

17 \pi/4 - 2\pi = \frac{9\pi}{4} -2\pi = \pi/4

For this case we know that sin (\pi/4) = \frac{\sqrt{2}}{2}

So then sin(\frac{17 \pi}{4}) = \frac{\sqrt{2}}{2}

b. cos (19pi / 6 )

We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to 2\pi:

19 \pi/6 - 2\pi = \frac{7\pi}{6}

For this case we know that cos (\pi/6) = \frac{\sqrt{3}}{2}

And we know that \frac{7\pi}{6} is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then cos(\frac{19 \pi}{6}) = -\frac{\sqrt{3}}{2}

c. tan(450pi)

For this case that any factor of \pi the sin function is equal to 0.

So from definition of tan we have this:

tan (450\pi) = \frac{sin(450 \pi)}{cos(450 \pi)}= \frac{0}{cos(450\pi)}= 0

4 0
3 years ago
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