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Bad White [126]
3 years ago
8

If -3 is a zero of f(x)= x^3+x^2-21x-45, find all of its zeros and their multiplicities

Mathematics
1 answer:
mr_godi [17]3 years ago
5 0

Answer: the zeros are -3, -3 and 5

Multiplicity is 2

Step-by-step explanation:

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Mrs James is a boutique owner who recently started selling Over-The-Knee boots. She buys them from her supplier for $300 per set
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Answer: $39.60

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How do you determine the quadratic equation having roots that are real numbers and equal​
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To determine the nature of roots of quadratic equations (in the form ax^2 + bx +c=0) , we need to calculate the discriminant, which is b^2 - 4 a c. When discriminant is greater than zero, the roots are unequal and real. When discriminant is equal to zero, the roots are equal and real.

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3 years ago
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Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0
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|x| = x for x ≥ 0

examples:

|3| = 3; |0.56| = 0.56; |102| = 102

|x| = -x for x < 0

examples:

|-3| = -(-3) = 3; |-0.56| = -(-0.56) = 0.56; |-102| = 102

--------------------------------------------------------------------------------

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

--------------------------------------------------------------------------------

h(x)=\dfrac{1}{3}|x-6|+4

Put the values of x to the equation of the function h(x):

x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6


6 0
2 years ago
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