Answer:
463833
expand
Medium
Solution
verified
Verified by Toppr
In △ABD and △ACD, we have
DB=DC ∣ Given
∠ADB=∠ADC ∣ since AD⊥BC
AD=AD ∣ Common
∴ by SAS criterion of congruence, we have.
△ABD≅△ACD
⇒AB=AC ∣ Since corresponding parts of congruent triangles are equal
Hence, △ ABC is isosceles.
Answer:
yes
Step-by-step explanation:
That's a huge number of points.
First: Calculate the value of 62 two pointers
62 * 2 = 124
Second: find the number of points that were due to 3 pointers.
151 - 124 = 27
Third: Divide by 3 to find the number of 3 pointers.
27÷3 = 9
Answer: There were 9 three pointers.
This construction called a pencil of lines. That figure can have 3,4,5.... any number of straights lines. The main rule is their common point. And it looks like spokes of a wheel. Here you can find a lot of vertical angles and rays.
The base case is the claim that
which reduces to
which is true.
Assume that the inequality holds for <em>n</em> = <em>k </em>; that
We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that
By the induction hypothesis,
Now compare this to the upper bound we seek:
because
in turn because
