<h3>
Answer: 80 degrees</h3>
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Explanation:
I'm assuming that segments AD and CD are tangents to the circle.
We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.
By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.
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Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.
Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.
We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.
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Here's what we have so far for quadrilateral DAEC
- angle A = 90
- angle E = 100
- angle C = 90
- angle D = unknown
Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees
A+E+C+D = 360
90+100+90+D = 360
D+280 = 360
D = 360-280
D = 80
Or a shortcut you can take is to realize that angles E and D are supplementary
E+D = 180
100+D = 180
D = 180-100
D = 80
This only works if AD and CD are tangents.
Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.
F(x)= -4x+9
Multiply f*x
xF=9+4x
Solve for X: -4x+xF=9+4x-4x=0
Combine like terms: -4x+4x=0
-4x+xF=9+0= 9
-9+-4x +xF = 0.
Answer:
yes they are
Step-by-step explanation: because 5x2=10 and 3x2=6 see they both multiply by 2
Let the three consecutive integers be x, x+1 and x+2.
According to the question, x+x+1+x+2=51
=>3x+3=51
=>3x+3-3=51-3
=>3x=48
=>x=16
Hence, first integer = 16,
second integer = 16 + 1 = 17 and
third integer = 16 + 2 = 18.
So you have set of these 6 numbers
And we know that we have 3 even numbers (2, 4, 6) and 3 odd numbers (1, 3, 5).
Now the possibility you will throw even number is 50%