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omeli [17]
3 years ago
11

Probability of numbers divisible by 9 from numbers 1-100?

Mathematics
1 answer:
quester [9]3 years ago
7 0
11
(9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99)
Those numbers divisible by 9 are the multiple of 9; thus need to know how many multiples of 9 there are between 1 and 100:

100 ÷ 9 = 11 r 1 ÷ last multiple of 9 is 11 × 9 (= 99)
→ There are 11 - 1 + 1 = 11 numbers between 1 and 100 which are divisible by 9.

(They are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.)
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Answer:

The nonzero value of c will be:

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Step-by-step explanation:

Given the function

f\left(x\right)=\:x^2-4x-c

f\left(c\right)=\:c^2-4c-c

as

f(c) = c

so

c=\:c^2-4c-c

switching the sides

c^2-4c-c=c

subtract c from both sides

c^2-4c-c-c=c-c

c^2-6c=0

c\left(c-6\right)=0

Using the zero factor principle

\:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

c=0\quad \mathrm{or}\quad \:c-6=0

so, the solutions to the quadratic equations are:

c=0,\:c=6

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5 0
3 years ago
36 − 9m^2 factorized
STatiana [176]

Answer:

Step-by-step explanation:

Note that the expression is the difference of squares.

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Answer:  Option 'E' is correct.

Step-by-step explanation:

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