36 equally-likely outcome: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1),(6,2), (6,3), (6,4), (6,5), (6,6)
Solution:
Outcomes with first number being old number and second being even number: (1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6) = 9 outcomes
P(old,even) = 9/36 =1/4 = 0.25
When x<span> approaches to </span><span>+∞</span><span> the function </span><span>e^<span>3x</span></span><span> becomes much bigger then </span><span>e^<span>−3x</span></span><span>, which obviously means that </span><span>e^<span>−3x</span></span><span> can be neglected in both numerator and denominator.
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Here's how I figured this out:
</span><span>lim <span>x →+∞ </span></span>= (<span><span><span>e^(<span>3x))</span></span>− (<span>e^(<span>−3x)) / (</span></span></span><span><span>e^<span>3x)) </span></span>+ (<span>e^(<span>−3x)) </span></span></span></span>= <span>lim <span>x → +∞ </span></span><span><span>e^<span>3x / </span></span><span>e^<span>3x </span></span></span>= 1
Answer:
N is M reflected across the x-axis; only the signs of the x-coordinates of M and N are different.
Step-by-step explanation:
The y-axis is vertical, and the x-axis is horizontal. If a point, or image, is reflected over the x-axis, in this case M and N, then the signs of the y-coordinates do not change, and instead the coordinates of the x-axis change.
A modulus is pretty much just the square root of x^2 + y^2, where y is the number that stands with the imaginary “i” and x is the real number. In this case, x is 2 and y is 3, so your modulus would be the square root of 4 + 9 (2^2 + 3^2), or the square root of 13.
