1+3=4
A counterexample is an a example that proves the statement false.
1+3 are not even numbers but they equal an even one, so it just proved the statement wrong.
I assume you're familiar with the limit
![\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto0%7D%5Cfrac%7B%5Csin%20ax%7D%7Bax%7D%3D1)
for
![a\neq0](https://tex.z-dn.net/?f=a%5Cneq0)
. We write the given limit in this form:
![\displaystyle\lim_{x\to0}\frac{\sin(-2x)}x=-2\cdot\lim_{x\to0}\frac{\sin(-2x)}{-2x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto0%7D%5Cfrac%7B%5Csin%28-2x%29%7Dx%3D-2%5Ccdot%5Clim_%7Bx%5Cto0%7D%5Cfrac%7B%5Csin%28-2x%29%7D%7B-2x%7D)
The limit on the RHS is 1, so we're left with -2.
His reasoning is not valid because they might actually appear to be equal to his sight, but should he measure them they might not exactly be.
So he just can't look at it and conclude. He can say it is equilateral if each angles measures 60 degrees.
Answer:
B
Step-by-step explanation:
540 is the sum of all angles in a pentagon
The ANGLE STP is 180-50 = 130
540 - 130 is 410
<h3><u>
Answer:</u></h3>
<em>Hence, the probability is:</em>
![\dfrac{1}{273}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B273%7D)
<h3><u>
Step-by-step explanation:</u></h3>
Each of four friends orders a sweatshirt from a catalog.
There are 15 colors of sweatshirts, 5 of which are all cotton and 10 of which are a blend.
We have to find the probability that the friend order only cotton sweatshirt:
As there are 5 cotton shirts out of which four are to be selected from a total of 15 coloured shirts.
Now we know that a probability is defined as:
Probability=( Number of favourable outcomes)/(Total number of outcomes)/( Total number of outcomes)
So, the probability is:
![Probability=\dfrac{5_C_4}{15_C_4}\\\\=\dfrac{\dfrac{5!}{4!\times 1!}}{\dfrac{15!}{4!\times 11!}}\\\\\\=\dfrac{1}{273}](https://tex.z-dn.net/?f=Probability%3D%5Cdfrac%7B5_C_4%7D%7B15_C_4%7D%5C%5C%5C%5C%3D%5Cdfrac%7B%5Cdfrac%7B5%21%7D%7B4%21%5Ctimes%201%21%7D%7D%7B%5Cdfrac%7B15%21%7D%7B4%21%5Ctimes%2011%21%7D%7D%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B273%7D)
Hence, the probability is:
![\dfrac{1}{273}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B273%7D)