Answer:
A. 70
Step-by-step explanation:
You can use the transversals and use vertical angles are congruent.
<u>Answer:</u>
- 6. 6.75
- 7. 27
- 8. 92/17
- 9. 49
<u>Step-by-step explanation:</u>
<u>- Question 6 -</u>
- 9/16 = x/12
- => 16x = 12 x 9
- => 16x = 108
- => x = 6.75
Hence, <u>the value of x in this proportion is 6.75.</u>
<h3>_______________________________</h3>
<u>- Question 7 -</u>
- -3 + x/18 = 12/9
- => -3/18 + x/18 = 4/3
- => x/18 = 4/3 + 1/6
- => x/18 = 8/6 + 1/6
- => x/18 = 9/6
- => x = 9/6 x 18
- => x = 27
Hence, <u>the value of x is 27</u>
<h3>_______________________________</h3>
<u>- Question 8 -</u>
17/15 = 10/2x - 2
=> 17(2x - 2) = 150
=> 34x - 34 = 150
=> 34x = 184
=> x = 184/34
=> x = 92/17
Hence, <u>the value of x is 92/17</u>
<h3>_______________________________</h3>
<u>- Question 9 -</u>
- x - 16/x + 6 = 3/5
- => 5(x - 16) = 3(x + 6)
- => 5x - 80 = 3x + 18
- => 2x = 98
- => x = 49
Hence, <u>the value of x is 49.</u>
<h3>_______________________________</h3>
Hoped this helped.
The equation of the newsletter function is C(x) = 75 + 0.25x and the function values are C(0) = 75, C(100) = 100, C(200) = 125 and C(300) = 150
<h3 /><h3>How to determine the newsletter function?</h3>
From the question, the given parameters are
Initial charge = $75.00
Rate per copy = $0.25 per copy
The equation of the newsletter function is then calculated as
Total = Initial charge + Rate per copy x Number of copies
Let x represents the number of copies
So, we have
Total = Initial charge + Rate per copy x x
This gives
C(x) = 75 + 0.25x
<h3>The function values for x = 0, 100, 200 and 300</h3>
When x = 0, we have
C(0) = 75 + 0.25 x 0 = 75
When x = 100, we have
C(100) = 75 + 0.25 x 100 = 100
When x = 200, we have
C(200) = 75 + 0.25 x 200 = 125
When x = 300, we have
C(300) = 75 + 0.25 x 300 = 150
Read more about linear equations at
brainly.com/question/4074386
#SPJ1
x = x
Consider x. Let x be a quantity denoted in the real numbers equal to x. Now, some properties of the real numbers include closure under the four basic operations.
