A. Alright, we want to multiply one equation by a constant to make it cancel out with the second. Since the first equation has a "blank" y, let's multiply the first equation by <em>2</em>.
3x-y=0 → 2(3x-y=0) = 6x - 2y = 0
5x+2y=22
The answer for this part would be: 6x - 2y = 0 and 5x + 2y = 22
B. So now we combine them:
6x - 2y = 0
+ + +
5x + 2y = 22
= = =
11x + 0 = 22 ← The answer
C. Now that we have the equation 11x = 22, we solve for x
11x = 22 ← Divide both sides by 11
x = 2 ← The answer
D. Now that we have x=2, we plug that back in to 5x+2y=22 and solve for y:
5(2)+2y = 22
10 + 2y = 22
2y = 12
y = 6
<u>Therefore, the solution to this problem is x = 2 and y = 6</u>
Answer:
Step-by-step explanation:
Let Q have coordinates x and y
<u>Using midpoint formula we find the values of x and y</u>
- (- 3 + x)/2 = 5 ⇒ -3 + x = 10 ⇒ x = 10 + 3 ⇒ x = 13
- (6 + y)/2 = 8 ⇒ 6 + y = 16 ⇒ y = 16 - 6 ⇒ y = 10
- Q = (13, 10)
Answer:
(a) 0
(b) f(x) = g(x)
(c) See below.
Step-by-step explanation:
Given rational function:

<u>Part (a)</u>
Factor the <u>numerator</u> and <u>denominator</u> of the given rational function:

Substitute x = -1 to find the limit:

Therefore:

<u>Part (b)</u>
From part (a), we can see that the simplified function f(x) is the same as the given function g(x). Therefore, f(x) = g(x).
<u>Part (c)</u>
As x = 1 is approached from the right side of 1, the numerator of the function is positive and approaches 2 whilst the denominator of the function is positive and gets smaller and smaller (approaching zero). Therefore, the quotient approaches infinity.

Answer:
hello,
answer C
Step-by-step explanation:
area of the rectangle=2b*b=2b²
area of the half cercle= 
Area in blue= 2b²-πc²/8
The answer would be 20,670