m∠PQR = m∠BQR + m∠PQB
We have m∠PQR = 116° and m∠BQR = 75°. Substitute:
116° = 75° + m∠PQB <em>subtract 75° from both sides</em>
41° = m∠PQB
<h3>Answer: m∠PQB = 41°</h3>
Answer:
-0.6
Step-by-step explanation:
66.5 square centimeters.
Start by separating the figure into a triangle and a rectangle.
We know the base of the triangle is 7 cm, so find the height by subtracting the total height of the figure by the height of the rectangle. 11 cm - 8 cm = 3 cm
Now, find the area of each shape. The area of a rectangle is length (base) times width (height). This is 7 cm * 8 cm, or 56 cm^2.
Then, find the area of the triangle. The area of a triangle is base times height times 1/2. So, it is 7 cm * 3 cm * 1/2.
21 cm^2 * 1/2
10.5 cm^2
Finally, add the areas together. 56 cm^2 + 10.5 cm^2 = 66.5 cm^2
Answer: center (0,0), vertices (+/-5,0), foci (+/-4.6,0)
Step-by-step explanation:
If the coefficient matrix has a pivot in each column, it means that it is shaped like this:
![A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Da_%7B1%2C1%7D%26a_%7B1%2C2%7D%26a_%7B1%2C3%7D%26a_%7B1%2C4%7D%5C%5C0%26a_%7B2%2C2%7D%26a_%7B2%2C3%7D%26a_%7B2%2C4%7D%5C%5C0%260%26a_%7B3%2C3%7D%26a_%7B3%2C4%7D%5C%5C0%260%260%26a_%7B4%2C4%7D%5Cend%7Barray%7D%5Cright%5D)
So, the correspondant system

will look like this:
![\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Da_%7B1%2C1%7D%26a_%7B1%2C2%7D%26a_%7B1%2C3%7D%26a_%7B1%2C4%7D%5C%5C0%26a_%7B2%2C2%7D%26a_%7B2%2C3%7D%26a_%7B2%2C4%7D%5C%5C0%260%26a_%7B3%2C3%7D%26a_%7B3%2C4%7D%5C%5C0%260%260%26a_%7B4%2C4%7D%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%5C%5Cx_2%5C%5Cx_3%5C%5Cx_4%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Db_1%5C%5Cb_2%5C%5Cb_3%5C%5Cb_4%5Cend%7Barray%7D%5Cright%5D)
This turn into the following system of equations:

The last equation is solvable for
: we easily have

Once the value for
is known, we can solve the third equation for
:

(recall that
is now known)
The pattern should be clear: you can use the last equation to solve for
. Once it is known, the third equation involves the only variable
. Once