It is given that line segment BC is congruent to line segment EC and that line segment AC is congruent to DC. Because of the vertical angles theorem, angle BCA is equal to angle DCE. Therefore, triangles CBA AND DEC are congruent by SAS. Using CPCTC, BA is equal to ED.
Answer:
Step-by-step explanation:
Answer:
740
Step-by-step explanation:
The n th term of an arithmetic series is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₃ = 7 and a₇ = (3 × 7) + 2 = 21 + 2 = 23 , then
a₁ + 2d = 7 → (1)
a₁ + 6d = 23 → (2)
Subtract (1) from (2) term by term
4d = 16 ( divide both sides by 4 )
d = 4
Substitute d = 4 into (1)
a₁ + 2(4) = 7
a₁ + 8 = 7 ( subtract 8 from both sides )
a₁ = - 1
The sum to n terms of an arithmetic series is
= 
[ 2a₁ + (n - 1)d ] , thus
= 
[ (2 × - 1) + (19 × 4) ]
= 10(- 2 + 76) = 10 × 74 = 740
Answer:
Claim 2
Step-by-step explanation:
The Inscribed Angle Theorem* tells you ...
... ∠RPQ = 1/2·∠ROQ
The multiplication property of equality tells you that multiplying both sides of this equation by 2 does not change the equality relationship.
... 2·∠RPQ = ∠ROQ
The symmetric property of equality says you can rearrange this to ...
... ∠ROQ = 2·∠RPQ . . . . the measure of ∠ROQ is twice the measure of ∠RPQ
_____
* You can prove the Inscribed Angle Theorem by drawing diameter POX and considering the relationship of angles XOQ and OPQ. The same consideration should be applied to angles XOR and OPR. In each case, you find the former is twice the latter, so the sum of angles XOR and XOQ will be twice the sum of angles OPR and OPQ. That is, angle ROQ is twice angle RPQ.
You can get to the required relationship by considering the sum of angles in a triangle and the sum of linear angles. As a shortcut, you can use the fact that an external angle is the sum of opposite internal angles of a triangle. Of course, triangles OPQ and OPR are both isosceles.
