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ale4655 [162]
3 years ago
10

6r-r+8(15-r)+23-6

Mathematics
1 answer:
IrinaVladis [17]3 years ago
4 0

Answer:

Therefore,

6r-r++(15-r)+23-6 = -3r+137 i.e

6r-r+8(15-r)+23-6    and

-3r+137 are Equivalent .

Step-by-step explanation:

To Check:

6r-r+8(15-r)+23-6 and

-3r+137 is Equivalent or Not

Solution:

Consider,

6r-r+8(15-r)+23-6  

Step 1 . Apply Distributive Property , A(B+C)=AB+AC we get

6r-r+8\times 15-8\times r+23-6  

6r-r+120-8r+23-6  

Step 2 . Combining Like Terms i.e r terms and the numbers we get

6r-r-8r+120+23-6  

-3r+137  

Which is Equivalent to the given expression

-3r+137

Therefore,

6r-r+8(15-r)+23-6= -3r+137   i.e

6r-r+8(15-r)+23-6      and

-3r+137                  ..........are Equivalent .

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Consider the following.
san4es73 [151]

Answer:

Anything in the form x = pi+k*pi, for any integer k

These are not removable discontinuities.

============================================================

Explanation:

Recall that tan(x) = sin(x)/cos(x).

The discontinuities occur whenever cos(x) is equal to zero.

Solving cos(x) = 0 will yield the locations when we have discontinuities.

This all applies to tan(x), but we want to work with tan(x/2) instead.

Simply replace x with x/2 and solve for x like so

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x/2 = arccos(0)

x/2 = (pi/2) + 2pi*k or x/2 = (-pi/2) + 2pi*k

x = pi + 4pi*k   or    x = -pi + 4pi*k

Where k is any integer.

If we make a table of some example k values, then we'll find that we could get the following outputs:

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  • x = 3pi
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and so on. These are the odd multiples of pi.

So we can effectively condense those x equations into the single equation x = pi+k*pi

That equation is the same as x = (k+1)pi

The graph is below. It shows we have jump discontinuities. These are <u>not</u> removable discontinuities (since we're not removing a single point).

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