17/21 and 5/6 as common denominators
1 answer:
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Answer:
15
Step-by-step explanation:
For these kind of problems, its best to use a Venn Diagram.
Lets break down the information:
<em>There are 50 students total.</em>
<em>25 take Math competiton classes.</em>
<em>29 take Geometry</em>
<em>12 take history and </em><em><u>no math classes</u></em>
<em>19 take </em><em><u>both math classes</u></em>
Now, make a Venn Diagram.
[Look at the photo attached. Also, the blank spaces in the Venn Diagram are ones that no student takes.]
The 12 taking history and no math classes and 3 kids not taking any of those classes look like they're the only ones not taking either math class.
12 + 3 = 15.
So the answer is 15.
[I'm not sure if this is correct or if I made a mistake so please let me know! Thank you! I hope this helps!]
:)
All the theorems are proved as follows.
<h3>What is a Triangle ?</h3>A triangle is a polygon with three sides , three vertices and three angles.
1. The Triangle sum Theorem
According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.
To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.
We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.
180° - ∠A + 180° - ∠B + 180° - ∠C = 360°
- ∠A - ∠B - ∠C = 360° - (180°+ 180°+ 180°) = -180°
∠A + ∠B + ∠C = 180°
(Hence Proved)
2. Isosceles Triangle Theorem
Considering an isosceles triangle ΔABC
with AB = AC, we have by sine rule;
as AB = AC
sin B = sin C
angle B = angle C
3.Converse of the Isosceles theorem
Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;
as ∠B= ∠C ,
AB = AC
4. Midsegment of a triangle theorem
It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.
Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC
In ΔABC and ΔADF
∠A ≅ ∠A
BA = 2 × DA, BC = 2 × FA
Hence;
ΔABC ~ ΔADF (SAS similarity)
BA/DA = BC/FA = DF/AC = 2
Hence AC = 2×DF
5.Concurrency of Medians Theorem
A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and the point of intersection is inside the triangle known as Centroid .
Consider a triangle ABC , X,Y and Z are the midpoints of the sides
Since the medians bisect the segment AB into AZ + ZB
BC into BX + XB
AC into AY + YC
Where:
AZ = ZB
BX = XB
AY = YC
AZ/ZB = BX/XB = AY/YC = 1
AZ/ZB × BX/XB × AY/YC = 1 and
the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).
To know more about Triangle
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