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
Answer:
V≈2412.74
Step-by-step explanation:
Answer:
256
Step-by-step explanation:
A calculator works well for this.
_____
None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.
1 to any power is still 1, so the factors (1^n) can be ignored.
After you ignore all of the things that can be ignored, your problem simplifies to ...
(2^2)(2^-3)^-2
The rules of exponents applicable to this are ...
(a^b)^c = a^(b·c)
(a^b)(a^c) = a^(b+c)
Then your product simplifies to ...
(2^2)(2^((-3)(-2)) = (2^2)(2^6)
= 2^(2+6)
= 2^8 = 256
Rational number, negative, integer