<u>Answer:</u>
So all the possible solutions are:
<u />
<u>Solution Steps:</u>
<em>First you need to solve the real inequality to understand how to find the rest of the possible equations. </em>
<u>Add 78 to both sides:</u>
- <u />
Cancels Out
<em>So now we know the real answer, but it ask for all possible answers. </em>
(Means anything larger than 300 when you plug it into x - 78 > 300.)
<u>Numbers that are greater than 378:</u>
1.)
(False)
2.)
(True)
3.)
(False)
4.)
(False)
5.)
(True)
6.)
(True)
7.)
(False)
8.)
(True)
______________________________

Answer:
-24/5
Step-by-step explanation:
The solution to the problem is as follows:
cscx < 0 means 1/sinx < 0, or sinx < 0.
sinx = -3/5 because you're dealing with a 3-4-5 triangle, and we established
sinx is negative.
sin2x = 2sinxcosx = 2*-3/5*4/5 = -24/25
cos(2x) = cos^2(x) - sin^2(x) = 16/25 - 9/25 = 5/25 = 1/5
tan2x = sin2x / cos2x = -24/5
Answer:
Emilio will have both activities again on the same day after 90 days from this Saturday.
Step-by-step explanation:
<em>To solve this question list the multiples of 30 and 9 to find the first common multiple because the two activities will happen again in </em><em>a number divisible by both 30 and 9.</em>
∵ The multiples of 30 are: 30, 60, 90, 120, ............
∵ The multiples of 9 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, .....
→ The common multiple is 90
∴ The common multiple of 30 and 9 is 90
→ That means the activities will meet again after 90 days from
this Saturday
Emilio will have both activities again on the same day after 90 days from this Saturday.
Step-by-step explanation:
you must have made a typo here.
none of the answer options are correct for the given problem.
let me just show you what you told me, and how I can solve at least this :
2a/4 = b/4
this simply means (multiply both sides by 4) :
2a = b
so, 5b is then (multiplying both sides by 5) :
5×2a = 5b
10a = 5b
but again, "10a" is not among the offered answers.
so, I don't know if you made a mistake with the basic problem or with the answer options.