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
That’s a lot of numbers lolll 3.14159265358979323846264338327950288419716939937510.... it goes on
Answer: 2343 / 256
Explanation
I will do this for you in two forms: 1) adding each term, and 2) using the general formula for the sum of geometric series.
1) Adding the terms:
4
∑ 3 (3/4)^i = 3 (3/4)^0 + 3 (3/4)^1 + 3 (3/4)^2 + 3 (3/4)^3 + 3 (3/4)^4
i=0
= 3 + 9/4 + 27/16 + 81/64 + 243/256 = [256*3 + 27*16 + 64*9 + 4*81 + 243] / 256 =
= 2343 / 256
2) Using the formula:
n-1
∑ A (r^i) = A [1 - r^(n) ] / [ 1 - r]
i=0
Here n - 1 = 4 => n = 5
r = 3/4
A = 3
Therefore the sum is 3 [ 1 - (3/4)^5 ] / [ 1 - (3/4) ] =
= 3 [ 1 - (3^5) / (4^5) ] / [ 1/4 ] = 3 { [ (4^5) - (3^5) ] / (4^5) } / {1/4} =
= (3 * 781) / (4^5) / (1/4) = 3 * 781 / (4^4) = 2343 / 256
So, no doubt, the answer is 2343 / 256
Answer:
Step 1: Flip the equation.
−x−2=y
Step 2: Add 2 to both sides.
−x−2+2=y+2
−x=y+2
Step 3: Divide both sides by -1.
−x
−1
=
y+2
−1
x=−y−2
Step-by-step explanation:
