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stiks02 [169]
2 years ago
14

Which of the following is the surface area of the right cylinder below?

Mathematics
2 answers:
Ierofanga [76]2 years ago
5 0

Answer:

The correct answer is A

Step-by-step explanation:

(Just took the test, A.pex)

muminat2 years ago
4 0

Answer:

The one you have selected is the right answer 126 times pi

Step-by-step explanation:

Exact answer - 395.840674

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Answer:

Mr.Daniels is a fajjett

Step-by-step explanation:

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il63 [147K]

Answer:

\frac{x-3}{2x-3}. hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

x^2-5x+6

Now let's find what ac is (it's just c since a=1...)

AC= 6

List factors of -6

\pm1, \pm2, \pm3, \pm6.

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

(x-2)(x-3)

Ok now let's factor the second equation:

2x^2-7x+6

Multiply a and c

AC = 12

List factors of 12:

\pm1, \pm2, \pm3, \pm4, \pm6, \pm12.

Factors that add up to -7 and multiply to 12:

-3\ and\ -4

Rewrite equation:

2x^2-4x-3x+6

Group terms:

(2x^2-4x)+(-3x+6)

Factor out GCF:

2x(x-2)-3(x-2)

Rewrite:

(2x-3)(x-2)

Now let's write out the equation using these factors:

\frac{(x-2)(x-3)}{(2x-3)(x-2)}.

Here we can factor out the x-2 and the simplified form is:

\frac{x-3}{2x-3}

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0
1 year ago
Nonpermissable replacement for a 4/9a?
My name is Ann [436]
If a=0, then the denominator is equal to 0.  Since you cannot divide by zero, you are not allowed to do this.
Hope This helps!

8 0
2 years ago
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