C. y - 3 = 2/3(x-3)
Nothing much to do here except examine each equation and plug in the numbers to see if it's true.
a. y + 3 = 3/2(x+3)
Try 3,3
3 + 3 = 3/2(3+3)
6 = 3/2(6). And no need to go further, it's obviously not equal.
b. y - 3 = 3/2(x-3)
Try 3,3
3 - 3 = 3/2(3-3)
0 = 3/2(0). OK. Let's try 6,5
5 - 3 = 3/2(6-3)
2 = 3/2(3)
2 = 9/2 And it's not true, so check the next one.
c. y - 3 = 2/3(x-3)
Try 3,3
3 - 3 = 2/3(3-3)
0 = 0. Check 6,5
5 - 3 = 2/3(6-3)
2 = 2/3(3)
2 = 2. Good. Both sample points work. This is the correct answer.
Just to be sure, let's check the next option
d. y + 3 = 2/3(x+3)
Try 3,3
3 + 3 = 2/3(3+3)
6 = 2/3(6). And doesn't match.
A. 2x2 – 4x – 6
Looking at the diagram, there two positive x's to the power of 2, and there are four negative x and six minus signs. Inspecting it and adding all the same terms, you will get the quadratic expression 2x^2 - 4x - 6.
Answer:
Option C
Step-by-step explanation:
![\sf log_w \ \dfrac{(x^{2}-6)^4}{\sqrt[3]{x^2+8} }=log_w \ (x^2-6)^4 - log_w(x^2+8)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csf%20log_w%20%5C%20%5Cdfrac%7B%28x%5E%7B2%7D-6%29%5E4%7D%7B%5Csqrt%5B3%5D%7Bx%5E2%2B8%7D%20%7D%3Dlog_w%20%5C%20%28x%5E2-6%29%5E4%20-%20log_w%28x%5E2%2B8%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
Two angles are supplementary if they add up to 180°.
?+ 30°= 180°
⇒ ?= 180° -30°
⇒ ?= 150°
The measurement of the second angle is 150°.
Answer:
The Proof for 
is below.
Step-by-step explanation:
To Prove:
Proof:
We know law of indices
Therefore,
Step 1: Applying law of indices
Step 2: Taking common 
we get
Step 3: 
so add 5 + 1 we get
Step 4: Cancel 6 from both denominator and numerator we get
...That is Divisible by 6 Proved
