Answer:i think its the first one
Step-by-step explanation:
Nothing else makes sense
Answer:
<h2>For c = 5 → two solutions</h2><h2>For c = -10 → no solutions</h2>
Step-by-step explanation:
We know

for any real value of <em>a</em>.
|a| = b > 0 - <em>two solutions: </em>a = b or a = -b
|a| = 0 - <em>one solution: a = 0</em>
|a| = b < 0 - <em>no solution</em>
<em />
|x + 6| - 4 = c
for c = 5:
|x + 6| - 4 = 5 <em>add 4 to both sides</em>
|x + 6| = 9 > 0 <em>TWO SOLUTIONS</em>
for c = -10
|x + 6| - 4 = -10 <em>add 4 to both sides</em>
|x + 6| = -6 < 0 <em>NO SOLUTIONS</em>
<em></em>
Calculate the solutions for c = 5:
|x + 6| = 9 ⇔ x + 6 = 9 or x + 6 = -9 <em>subtract 6 from both sides</em>
x = 3 or x = -15
If you go on my profile you will see a similar problem that I already answered today.
Lets start by using a formula (don't remember the name)
(y-y1)=M(x-x1)
M is slope and y is first y value and y1 is second y value same applies to X.
Substitute in the values.
(5-q) = 10(-6-(-7))
5-q = 10*(1)
5-q = 10
-q = 5
q = -5
Check:
Substitute in the value of "q" and use the formula to find slope:
(y1-y)/(x1-x)
Substitute
(-5-5)/(-7-(-6))
-10/-1
10
Your Answer:
The value of "q" is -5
change into improper fraction then do the problem then turn back into mixed number
Answer:
I swear I hate those people who put links