What is the value of W in a exterior angle theorem
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Answer:
The factored form of given polynomial is
Step-by-step explanation:
Given the polynomial
We have to factor the above polynomial completely.
Take common from each term.
Here, the quadratic polynomial is prime i.e it cannot be factored into lower degree polynomial. It has complex roots.
Hence, the factored form of given polynomial is
The matrix equation that represents this situation is
![\left[\begin{array}{ccc}3&2&0\\1&0&4\\3&1&1\end{array}\right]*\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}13.50\\16.50\\14.00\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%260%5C%5C1%260%264%5C%5C3%261%261%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D13.50%5C%5C16.50%5C%5C14.00%5Cend%7Barray%7D%5Cright%5D%20)
Use technology to find the inverse of matrix A:
![A^{-1}= \left[\begin{array}{ccc}-\frac{2}{5}&-\frac{1}{5}&\frac{4}{5}\\ \frac{11}{10}&\frac{3}{10}&-\frac{6}{5}\\\frac{1}{10}&\frac{3}{10}&-\frac{1}{5}\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B2%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%26%5Cfrac%7B4%7D%7B5%7D%5C%5C%0A%5Cfrac%7B11%7D%7B10%7D%26%5Cfrac%7B3%7D%7B10%7D%26-%5Cfrac%7B6%7D%7B5%7D%5C%5C%5Cfrac%7B1%7D%7B10%7D%26%5Cfrac%7B3%7D%7B10%7D%26-%5Cfrac%7B1%7D%7B5%7D%5Cend%7Barray%7D%5Cright%5D%20)
Multiplying A inverse by B, we get the solution matrix
[tex]\left[\begin{array}{ccc}2.50\\3\\3.50\end{array}\right][\tex]
Use technology to find the inverse of matrix A:
Multiplying A inverse by B, we get the solution matrix
[tex]\left[\begin{array}{ccc}2.50\\3\\3.50\end{array}\right][\tex]
It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
</span><span>
</span><span>In notation we write respectively
</span>
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²
That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers