In ∆FDH, there are two slash marks in two of its legs. This indicates that this triangle is isosceles. If a triangle is isosceles, then it will have two congruent sides and therefore have two congruent angles.
In ∆FDH, angle D is already given to us as the measure of 80°. We can find out the measure of the other angles of this triangle by using the equation:
80 + 2x = 180
Subtract 80 from both sides of the equation.
2x = 100
Divide both sides by 2.
x = 50
This means that angle F and angle H in ∆FDH both measure 50°.
Now, moving over to the next smaller triangle in the picture is ∆DHG. In this triangle, there are also two legs that are congruent which once again indicates that this triangle is isosceles.
First, we have to solve for angle DHG and we do that by using the information obtained from solving for the angles of the other triangle.
**In geometry, remember that two or more consecutive angles that form a line will always be supplementary; the angles add up to 180°.**
In this case angle DHF and angle DHG are consecutive angles which form a linear pair. So, we can use the equation:
Angle DHF + Angle DHG = 180°
50° + Angle DHG = 180°.
Angle DHG = 130°.
Now that we know the measure of one angle in ∆DHG, we can use the same method as the previous step for solving the missing angles. Use the equation:
130 + 2x = 180
2x = 50
x = 25
The other two missing angles of ∆DHG are 25°. This means that the measure of angle 1 is also 25°.
Solution: 25°
Answer:
Step-by-step explanation:
To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:
![A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%262%262%5C%5C-7%26-3%265%26-8%5C%5C4%261%261%261%5C%5C3%267%26-1%261%5Cend%7Barray%7D%5Cright%5D)
And the vector B is formed with the solution of each equation of the system:![b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]](https://tex.z-dn.net/?f=b%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%5C%5C-3%5C%5C6%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called 
.
![A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]](https://tex.z-dn.net/?f=A_%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%262%262%5C%5C-7%26-3%265%26-8%5C%5C4%266%261%261%5C%5C3%261%26-1%261%5Cend%7Barray%7D%5Cright%5D)
The value of y using Cramer's rule is:
Find the value of the determinant of each matrix, and divide:
the difference of MA is theres only A
the difference is the order
yes AB can be written as BA
Answer: v = -5/8
Step-by-step explanation:
Answer:
Step-by-step explanation:
