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
Step-by-step explanation:
Enter the following numbers in the boxes provided sequentially.
First box : 7
Second box: 2
Third box: 3
Fourth box : 4
Fifth box: 8
Sixth box: 3
Answer:
Parallelograms I, II, and IV
Step-by-step:
Area of parallelograms:
I. A=3*5=15 units squared
II. A=5*3=15 units squared
III. A=4*4=16 units squared
IV. A=5*3=15 units squared
So, parallelograms I, II, and IV have the same area of 15 units squared.
Answer:58 approximately c
Step-by-step explanation:
