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crimeas [40]
3 years ago
11

HELP NEEDED !!! PLEASE

Mathematics
1 answer:
alexgriva [62]3 years ago
4 0

The external angles of a 12-gon add up to 360 degrees so each is 30 degrees.  That makes the internal angle the supplement, PQR=150 degrees.

We have an isosceles triangle so the other two angles add to 30 degrees, so 15 degrees.

Answer: 15 degrees

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Determine the approximate circumference of a circle with diameter 18 units.
Sphinxa [80]

Answer: 56.549 or 56.5

Explanation:

C= (pie) 3.14 * d = 2 * 3.14 * r

D= 18

C = 3.14 * 18 = 18 * 3.14

C = 56.549 or 56.5

8 0
2 years ago
If sin theta = cos theta then find the value of theta
Ksenya-84 [330]

Step-by-step explanation:

\because \sin  \theta =  \cos  \theta  \\   \\ \therefore \: \sin  \theta =  \sin(90 \degree -   \theta)  \\   \\ \therefore \:  \theta =  90 \degree -   \theta \\   \\ \therefore \:   \theta + \theta=  90 \degree \\  \\ \therefore \:  2\theta =  90 \degree \\  \\\therefore \:  \theta =   \frac{90 \degree }{2} \\  \\ \huge \orange{\boxed{\therefore \:   \theta =   45 \degree}}\\  \\

4 0
3 years ago
Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)
vfiekz [6]

Answer:

Step-by-step explanation:

given a point (x_0,y_0) the equation of a line with slope m that passes through the  given point is

y-y_0 = m(x-x_0) or equivalently

y = mx+(y_0-mx_0).

Recall that a line of the form y=mx+b, the y intercept is b and the x intercept is \frac{-b}{m}.

So, in our case, the y intercept is (y_0-mx_0) and the x  intercept is \frac{mx_0-y_0}{m}.

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph (x_0,\frac{1}{x_0}). Which means that y_0=\frac{1}{x_0}

The slope of the tangent line is given by the derivative of the function evaluated at x_0. Using the properties of derivatives, we get

y' = \frac{-1}{x^2}. So evaluated at x_0 we get m = \frac{-1}{x_0^2}

Replacing the values in our previous findings we get that the y intercept is

(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}

The x intercept is

\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0

The triangle in consideration has height \frac{2}{x_0} and base 2x_0. So the area is

\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0
3 years ago
The difference of twice a number and 5 is 3.
Elan Coil [88]
Option C is the right way to solve the equation.
6 0
3 years ago
WILL GIVE BRANLIEST!!! Pls help! Determine the coordinates of the point on the straight line y=3x+1 that is equidistant from the
iren [92.7K]

Let , coordinate of points are P( h,k ).

Also , k = 3h + 1

Distance of P from origin :

d=\sqrt{h^2+k^2}

Distance of P from ( -3, 4 ) :

d=\sqrt{(h+3)^2+(k-4)^2}

Now , these distance are equal :

h^2+(3h+1)^2=(h+3)^2+(3h+1-4)^2\\\\h^2+(3h+1)^2=(h+3)^2+(3h-3)^2

Solving above equation , we get :

P=(\dfrac{16}{21},\dfrac{23}{7})

Hence , this is the required solution.

6 0
3 years ago
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