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Lina20 [59]
3 years ago
15

I'm having trouble with this problem, can someone please help me?

Mathematics
1 answer:
Anastasy [175]3 years ago
4 0
The formula you use is 
V = (4/3)*pi*r^3

In this case, r = 6 is the radius, so the volume V is
V = (4/3)*pi*r^3
V = (4/3)*pi*6^3
V = (4/3)*pi*216
V = (4/3)*216*pi
V = 288pi

Answer: Upper Right-hand Choice
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I need help with this
N76 [4]

Answer:

The x-intercepts are x = 1 , x = 2 , x = 3

The y-intercept is -6

Step-by-step explanation:

* Lets explain how to solve the problem

- To find the x-intercept of a function substitute f(x) by 0

- To find the y-intercept of a function substitute x by 0

- To find the factors of quadratic function use the long division

* Lets solve the problem

∵ f(x) = x³ - 6x² + 11x - 6

∵ (x - 3) is one of its factors

- Use the long division to find the other factors

∵ x³ - 6x² + 11x - 6 ⇒ dividend

∵ x - 3 ⇒ divisor

# Divide the 1st term in the dividend by the 1st term of the divisor

∵ x³ ÷ x = x²

# Multiply x² by the divisor (x - 3)

∵ x²(x - 3) = x³ - 3x²

# subtract it from the dividend

∵ (x³ - 6x² + 11x - 6) - (x³ - 3x²) = (x³ - x³) + (-6x² + 3x²) +11x - 6

∴ The dividend is -3x² + 11x - 6

# Divide the 1st term in the dividend by the 1st term of the divisor

∵ -3x² ÷ x = -3x

# Multiply -3x by the divisor (x - 3)

∴ -3x(x - 3) = -3x² + 9x

# subtract it from the dividend

∵ (-3x² + 11x - 6) - (-3x² + 9x) = (-3x² - 3x²) + (11x - 9x) - 6

∴ The dividend is 2x - 6

# Divide the 1st term in the dividend by the 1st term of the divisor

∵ 2x ÷ x = 2

# Multiply 2 by the divisor (x - 3)

∴ 2(x - 3) = 2x - 6

# subtract it from the dividend

∴ (2x - 6) - (2x - 6) = (2x - 2x) + (-6 + 6) = 0

∴ (x³ - 6x² + 11x - 6) ÷ (x - 3) = x² - 3x + 2

∴ The factors of f(x) are (x - 3)( x² - 3x + 2)

- The factor (x² - 3x + 2) can factorize into two bracket

∵ The last term is positive and the middle term is negative than the

   two brackets have the middle sign (-)

∵ x × x = x² ⇒ 1st terms in the two brackets

∵ 2 × 1 = 2 ⇒ 2nd terms in the two brackets

∵ 2 × x = 2x

∵ 1 × x = x

∵ 2x + x = 3x

∴  (x² - 3x + 2) = (x - 2)(x - 1)

∴ f(x) = (x - 3)(x - 2)(x - 1)

- To find the x-intercept put f(x) = 0

∴ (x - 3)(x - 2)(x - 1) = 0

- That means each bracket = 0

∵ x - 3 = 0 ⇒ add 3 for both sides

∴ x = 3

∵ x - 2 = 0 ⇒ add 2 for both sides

∴ x = 2

∵ x - 1 = 0 ⇒ add 1 for both sides

∴ x = 1

∴ The x-intercepts are x = 1 , x = 2 , x = 3

- To find the y-intercept put x = 0

∵ f(x) = x³ - 6x² + 11x - 6

∵ x = 0

∴ f(0) = 0 - 0 + 0 - 6 = -6

∴ The y-intercept is -6

4 0
3 years ago
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