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3 years ago

(will give brainliest) find the value of x

Mathematics

2 answers:

Sedaia [141]3 years ago

5 0

Answer:

x = 180 - [(180 - 3x) + (180 - 2x)]

Step-by-step explanation:

Start off by finding the angles of the triangle

Angle F = 180 - 3x

The angle across from I (which I will call I) = 180 - 2x

Angle G = 180 - (F + I)

Now that we know what G is, we know what x is because the Alternate Exterior Angles Theorem states that if a pair of parallel lines are cut by a transversal, then the alternate exterior angles are congruent. So pretty much X = G

Therefore x = 180 - (F + I) or otherwise said as:

x = 180 - [(180 - 3x) + (180 - 2x)]

I hope this is helpful :)

ikadub [295]3 years ago

3 0

Answer: x=45
Explanation:

(180-2x)+(180-3x)+x=180 (angle sum of triangle)
180-2x+180-3x+x=180
180+180-2x-3x+x=180
360-4x=180
360-4x-360=180-360
-4x=(-180)
(-4x)/(-4)=(-180)/(-4)
x=45
Hope this will help.

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kirill [66]

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3 years ago

Which are the solutions of x2 = 19x + 1?

Finger [1]

Answer:

$(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})$

Step-by-step explanation:

we have

$x^2=19x+1$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2}-19x-1=0$

$a=1\\b=-19\\c=-1$

substitute in the formula

$x=\frac{-(-19)\pm\sqrt{-19^{2}-4(1)(-1)}} {2(1)}$

$x=\frac{19\pm\sqrt{365}} {2}$

$x=\frac{19+\sqrt{365}} {2}$

$x=\frac{19-\sqrt{365}} {2}$

$(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})$

therefore

StartFraction 19 minus StartRoot 365 EndRoot Over 2 EndFraction comma StartFraction 19 + StartRoot 365 EndRoot Over 2 EndFraction

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3 years ago

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Brums [2.3K]

Answer: sqrt(2)/2 which is choice D

======================================================

Explanation

(3pi/4) radians converts to 135 degrees after multiplying by the conversion factor (180/pi).

The angle 135 degrees is in quadrant 2. We subtract the angle 135 from 180 to find the reference angle

180-135 = 45

Then you can use a 45-45-90 triangle to determine that the ratio of opposite over hypotenuse is sqrt(2)/2

sine is positive in quadrant 2

------------

Alternatively, you can use a unit circle. Refer to the diagram below. In red, I've circled the angle 3pi/4 radians. The terminal point for this angle has a y coordinate of sqrt(2)/2

Recall that y = sin(theta).

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