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kiruha [24]
2 years ago
8

Four sevenths plus what equals six sevenths

%20%5Cfrac%7B4%7D%7B7%7D%20%20" id="TexFormula1" title=" \frac{6}{7} = m + \frac{4}{7} " alt=" \frac{6}{7} = m + \frac{4}{7} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
ser-zykov [4K]2 years ago
3 0

M= 2/7. I am not sure if it will put it in fraction form, but M=two sevenths.

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Sin(x+22)°=cos(2x−7)°
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To answer this, it must be noted that the trigonometric functions sine and cosine will have the same values if the angles are complementary (meaning, their sum is 90°). 
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The value of x from the equation is 25. 
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9514 1404 393

Answer:

  (x +1)^2 +(y -3)^2 = 4

Step-by-step explanation:

The equation of a circle centered at (h, k) with radius r is ...

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The graphed circle is centered at (h, k) = (-1, 3), and has radius 2, so its equation is ...

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Please help me with these
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When we approach limits, we are finding values that are infinitesimally approaching this x-value. Essentially, we consider the approximate location that this root or limit appears. This is essential when it comes to taking Calculus, and finding the limit or rate of change of a function.

When we are attempting limits questions, there are several tests we attempt first.

1. Evaluate the limit by substituting the value of the x-value as it approaches the value (direct evaluation of a limit)
2. Rearrangement of the function, such that we can evaluate the limit.
3. (TRIGONOMETRIC PROPERTIES)
\lim_{x \to 0} (\frac{sinx}{x}) = 1
\lim_{x \to 0} (\frac{tanx}{x}) = 1
4. Using L'Hopital's Rule for indeterminate limits, such as 0/0, -infinity/infinity, or infinity/infinity.

For example:

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We can do this using the first and second method.
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Substitute x = 0 to the function.
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= \frac{1}{5}

<em>Method 2: Rearranging the function
</em>

We can see that x - 25 can be rewritten as: (√x - 5)(√x + 5)
By rewriting it in this form, the top will cancel with the bottom easily, and our limit comes out the same.

\lim_{x \to 0}\frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5)(\sqrt{x} + 5)}
= \lim_{x \to 0}\frac{1}{(\sqrt{x} + 5)}}
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Every example works exactly the same way, and by remembering these criteria, every limit question should come out pretty naturally.
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