Segments AB and CD bisect each other at point O. Prove that ∆ACD is congruent to ∆BDC.
1 answer:
Answer:
Step-by-step explanation:
TO prove this we have to prove one of following.
SSS ( all three sides are equal )
SAS ( two sides and their middle angle is equal )
ASA ( two angles and their middle side are equal )
I am proving both triangles equal by ASA method.
The complete proof and explanation is given in attachment below.
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15. The vertex form of f(x) = x² + 8x - 10 is f(x) = (x + 4)² - 26
17. The standard form of g(x) = (x + 3)² - 4 is g(x) = x² + 6x + 5
Step-by-step explanation:
Any quadratic function has:
1. A vertex form form f(x) = a(x - h)² + k, where (h , k) are the
coordinates of its vertex point
2. A standard form g(x) = ax² + bx + c, where a , b , c are constant and
a is the coefficient of x², b is the coefficient of x and c is the
y-intercept
3. the x-coordinate of the vertex point h = and
k = f(h)
15.
∵ f(x) = x² + 8x - 10
∴ a = 1 and b = 8
∵ h =
∴ h =
∴ h = -4
∵ k = f(h)
∴ k = f(-4)
- Substitute x by -4 in f(x)
∴ k = (-4)² + 8(-4) - 10
∴ k = 16 - 32 - 10
∴ k = -26
∴ f(x) = (x - -4)² + (-26)
∴ f(x) = (x + 4)² - 26
The vertex form of f(x) = x² + 8x - 10 is f(x) = (x + 4)² - 26
17.
∵ g(x) = (x + 3)² - 4
- Lets solve the power 2 of the bracket:
(1st + 2nd)² = (1st)(1st) + 2(1st)(2nd) + (2nd)(2nd)
∵ (x + 3)² = (x)(x) + 2(x)(3) + (3)(3)
∴ (x + 3)² = x² + 6x + 9
∴ (x + 3)² - 4 = (x² + 6x + 9) - 4
∵ 9 - 4 = 5
∴ (x + 3)² - 4 = x² + 6x + 5
The standard form of g(x) = (x + 3)² - 4 is g(x) = x² + 6x + 5
Learn more:
You can learn more about the vertex form of quadratic function in brainly.com/question/9390381
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