Answer:
3040cm
Step-by-step explanation:
you simply multiply 10 by 19 to get 190 than do 190 times 16 where u would go 6 times 0 =0 6 times 9 =54 leave the 4 carry the 5 than 6 times 1 is 6 but add the 5 to get 1140 than for the 1 simply put a placeholder 0 than just put 190 since you're multiplying by one to get 3040
Definitely the question is incomplete
Using for loop in java to print positive integers ,let say first 50 integers
So the loop is:
int i ;
for( i= 0; i<= 50; i++ ) {
print (i);
}
The looping will going to start from the 1 and stop to 50
Answer:
<h3>#1</h3>
The normal overlaps with the diameter, so it passes through the center.
<u>Let's find the center of the circle:</u>
- x² + y² + 2gx + 2fy + c = 0
- (x + g)² + (y + f)² = c + g² + f²
<u>The center is:</u>
<u>Since the line passes through (-g, -f) the equation of the line becomes:</u>
- p(-g) + p(-f) + r = 0
- r = p(g + f)
This is the required condition
<h3>#2</h3>
Rewrite equations and find centers and radius of both circles.
<u>Circle 1</u>
- x² + y² + 2ax + c² = 0
- (x + a)² + y² = a² - c²
- The center is (-a, 0) and radius is √(a² - c²)
<u>Circle 2</u>
- x² + y² + 2by + c² = 0
- x² + (y + b)² = b² - c²
- The center is (0, -b) and radius is √(b² - c²)
<u>The distance between two centers is same as sum of the radius of them:</u>
<u>Sum of radiuses:</u>
<u>Since they are same we have:</u>
- √(a² + b²) = √(a² - c²) + √(b² - c²)
<u>Square both sides:</u>
- a² + b² = a² - c² + b² - c² + 2√(a² - c²)(b² - c²)
- 2c² = 2√(a² - c²)(b² - c²)
<u>Square both sides:</u>
- c⁴ = (a² - c²)(b² - c²)
- c⁴ = a²b² - a²c² - b²c² + c⁴
- a²c² + b²c² = a²b²
<u>Divide both sides by a²b²c²:</u>
Proved
