Answer:
Step-by-step explanation:
<span>1. 3a²-a = a(3a - 1)
2. 7ab³-14b = 7b(ab^2 - 2)
3. x³-x²+x = x(x^2 - x + 1)
4. 12x³-xy² = x(12x^2 - y^2)
5. 5x-6x²+7x³ = x(5 - 6x + 7x^2)
6. 7a+8ab+9a² = a(7 + 8b + 9a)
7.3x²-3xy+6xy²
</span>= 3x(x - y + 2y^2)
Answer:
UT = 104
∠R = 126°
Step-by-step explanation:
Part 1: Finding UT
The symbols on the triangles indicate that the triangles have the same side lengths.
That means 2x + 84 = 14x - 36
We can find the length of UT by solving for x
2x+84=14x−36
<u>Step 1: Subtract 14x from both sides.</u>
2x + 84 − 14x = 14x − 36 − 14x
−12x + 84 = −36
<u>Step 2: Subtract 84 from both sides.</u>
−12x + 84 − 84 = −36 − 84
−12x = −120
<u>Step 3: Divide both sides by -12.</u>
-12x / -12 = -120 / -12
x = 10
Now we know x=10, we can substitute 10 for x to get UT
UT = 2x + 84
UT = 2(10) + 84
UT = 20 + 84
UT = 104
So the length of UT is 104
Part 2: Finding ∠R
Since we know angle R is equal to angle U, we know
10y - 14 = 5y + 56
We can solve for y to find R
<u>Step 1: Subtract 5y from both sides.</u>
10y − 14 − 5y = 5y + 56 − 5y
5y − 14 = 56
<u>Step 2: Add 14 to both sides.</u>
5y−14+14=56+14
5y=70
<u>Step 3: Divide both sides by 5.</u>
5y/5 = 70/5
y=14
Now that we know y=14, we can substitute that value to find ∠R
∠R = 10y - 14
∠R = 10(14) - 14
∠R = 140 - 14
∠R = 126°
Answer:
![\left[\begin{array}{cc}-\frac{1}{2}&\frac{1}{6}\\-\frac{1}{2}&\frac{1}{3}\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B1%7D%7B6%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B1%7D%7B3%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
If a 2x2 matrix is given as:
![\left[\begin{array}{cc}a&b\\c&d\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5C%5C%5Cend%7Barray%7D%5Cright%5D)
The inverse is:
![\frac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bad-bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dd%26-b%5C%5C-c%26a%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Our matrix given is:
![\left[\begin{array}{cc}-4&2\\-6&6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-4%262%5C%5C-6%266%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<em />
<em>Using the formula, let's find the inverse:</em>
<em>![\frac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\\\end{array}\right]\\=\frac{1}{(-4)(6)-(-6)(2)}\left[\begin{array}{cc}6&-2\\6&-4\\\end{array}\right]\\=\frac{1}{-12}\left[\begin{array}{cc}6&-2\\6&-4\\\end{array}\right]\\=\left[\begin{array}{cc}-\frac{1}{2}&\frac{1}{6}\\-\frac{1}{2}&\frac{1}{3}\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bad-bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dd%26-b%5C%5C-c%26a%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%5Cfrac%7B1%7D%7B%28-4%29%286%29-%28-6%29%282%29%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-2%5C%5C6%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%5Cfrac%7B1%7D%7B-12%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-2%5C%5C6%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B1%7D%7B6%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B1%7D%7B3%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
</em>
