From A draw the altitude AH intersecting BC in H
Let's prove that triangle ABH is congruent to triangle ACH
The above 2 triangles are right triangles due to the altitude AH
Angle B= Angle C (given)
Angle AHB = Angle AHC =90° (since AH is the altitude)
Then angle BAH = CAH (both complementary to B & C respectively
And AH is a common side
Now Tri. ABH = Tri. ACH because ASA, hence AB=AC
9 and 3/8 is 75/8 as a mixed #.
Answer:
He's wrong.
Step-by-step explanation:
30% of 50 is 3/10 * 50 = 15
50% of 30 is 1/2 * 30 = 15
As you can see, 30% of 50 is the same as 50% of 30.
Answer:
(-15/13, 0)
Step-by-step explanation:
For x-intercept, ALWAYS put y=0, and vice versa is true.
Given equation: 13x+y=-15
X-intercept (Put y=0);
13x + (0) = -15
13x=-15
x=-15/13
Hence x=-15/13 and y=0 (for x-intercept)
In co-ordinate form, we will write it as:
(x,y) = (-15/13, 0)
Its a quadratic equation which is flipped across the x-axis but there will be no change in the domain as it will remain:
Domain : Set of all real numbers.
