Using the speed - distance relationship, the time left before the appointed time is 27 minutes.
<u>Recall</u><u> </u><u>:</u>
<u>At</u><u> </u><u>10mph</u><u> </u><u>:</u>
- Distance = 10 × (t + 3) = 10t + 30 - - - (1)
<u>At</u><u> </u><u>12</u><u> </u><u>mph</u><u> </u><u>:</u>
- Distance = 12 × (t - 2) = 12t - 24 - - - - (2)
<em>Equate</em><em> </em><em>(</em><em>1</em><em>)</em><em> </em><em>and</em><em> </em><em>(</em><em>2</em><em>)</em><em> </em><em>:</em>
10t + 30 = 12t - 24
<em>Collect</em><em> </em><em>like</em><em> </em><em>terms</em><em> </em>
10t - 12t = - 24 - 30
-2t = - 54
<em>Divide</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>by</em><em> </em><em>-</em><em> </em><em>2</em>
t = 54 / 2
t = 27
Hence, the time left before the appointed time is 27 minutes.
Learn more : brainly.com/question/25669152
the answer I believe is 0.63. I did something similar awhile ago
It would be 15 because the 35 people at most minus the 20 students equals 15 so that would be the greatest number of adults allowed to be there.
It would be
(X^4+2X^2) -119(X^2-2X) then you would simplify
X^4+2X^2-119X^2-2X
So it would be
X^4+117X^2-2X=0
So basically X is undefined
The first and crucial thing we want to take notice is that the lines DF and EG intersect at point H which creates 4 different triangles in the rhombus. The innermost angles of DGH and EFH are vertical angles and vertical angles are congruent. So if the angles of the triangles are congruent than the triangles themselves are congruent. This is supported by the vertical angles theorem.
