Answer: 9a-9b-6
7(3a−2b)+5b−3(4a+2)
Distribute:
7(3a−2b)+5b−3(4a+2)
=(7)(3a)+(7)(−2b)+5b+(−3)(4a)+(−3)(2)
=21a+−14b+5b+−12a+−6
Combine Like Terms:
=21a+−14b+5b+−12a+−6
=(21a+−12a)+(−14b+5b)+(−6)
=9a+−9b+−6
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
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