2. 1 1/2 each
3. 1 1/4 each
4. 1 3/4 each
5. 3/4 each
Answer:
4±√5
EXPLAINED ANSWER:
Since the variable is a binomial and it is squared, we apply square root on both side, and apply the ± sign on the right side (since the solution of a square root can be both positive and negative), and later pass the 4 positive to the other side.
√25 can be (5)(5) or (-5)(-5)
Answer:
1. D
2. B
3. A
Step-by-step explanation:
Question 1:
The pair of <JKL and <LKM can be referred to as linear pairs. They are two adjacent angles that are formed from the intersecting of two lines.
Question 2:
Given that <KLM = x°
<KML = 50°
<JKL = (2x - 15)°
According to the exterior angle theorem, exterior ∠ JKL = <KLM + KML.
2x - 15 = x + 50
Solve for x
2x - x = 15 + 50
x = 65
Therefore, <KLM = 65°
QUESTION 3:
<JKL = 2x - 15
Plug in the value of x
<JKL = 2(65) - 15
= 130 - 15
<JKL = 115°
Answer:
A)11
Step-by-step explanation:
These are matrices one dimensional with one column and 3 rows each.
-The product of the matrices is obtained by multiplying the correspond values and summing up;
![pq=\left[\begin{array}{ccc}3\\2\\-1\end{array}\right] \times\left[\begin{array}{ccc}5\\-1\\2\end{array}\right] \\\\\\\\=(3\times 5)+(2\times -1)+(-1\times 2)\\\\=15+-2+-2\\\\=11](https://tex.z-dn.net/?f=pq%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C2%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C-1%5C%5C2%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5C%5C%3D%283%5Ctimes%205%29%2B%282%5Ctimes%20-1%29%2B%28-1%5Ctimes%202%29%5C%5C%5C%5C%3D15%2B-2%2B-2%5C%5C%5C%5C%3D11)
Hence, the product of p and q is 11
Y = 3 because in a rectangle, the diagonals are ALWAYS congruent. so you would have:
5y - 2 = 4y + 1
-4y+2 -4y + 2
y = 3