<span>-7 x (-9) = +63
because if we multiply negative by negative we obtain +</span>
I suppose you just have to simplify this expression.
(2ˣ⁺² - 2ˣ⁺³) / (2ˣ⁺¹ - 2ˣ⁺²)
Divide through every term by the lowest power of 2, which would be <em>x</em> + 1 :
… = (2ˣ⁺²/2ˣ⁺¹ - 2ˣ⁺³/2ˣ⁺¹) / (2ˣ⁺¹/2ˣ⁺¹ - 2ˣ⁺²/2ˣ⁺¹)
Recall that <em>n</em>ª / <em>n</em>ᵇ = <em>n</em>ª⁻ᵇ, so that we have
… = (2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺³⁾ ⁻ ⁽ˣ⁺¹⁾) / (2⁽ˣ⁺¹⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾)
… = (2¹ - 2²) / (2⁰ - 2¹)
… = (2 - 4) / (1 - 2)
… = (-2) / (-1)
… = 2
Another way to get the same result: rewrite every term as a multiple of <em>y</em> = 2ˣ :
… = (2²×2ˣ - 2³×2ˣ) / (2×2ˣ - 2²×2ˣ)
… = (4×2ˣ - 8×2ˣ) / (2×2ˣ - 4×2ˣ)
… = (4<em>y</em> - 8<em>y</em>) / (2<em>y</em> - 4<em>y</em>)
… = (-4<em>y</em>) / (-2<em>y</em>)
… = 2
-0.10416 /.going on forever
The quotient to p/q = -3 is 4
Answer:
Colin has <em>8 sheets </em>left for his third class.
Step-by-step explanation:
Given that:
Total Number of pieces of papers = 
Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad
Writing the equation:
Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.
Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class
So, the answer is:
Colin has <em>8</em> <em>sheets </em>left for his third class.
