Answer:
Yes.
Here’s an easy (and cheap) one: ![\\\\\sqrt 2\\-\sqrt 2](https://tex.z-dn.net/?f=%5C%5C%5C%5C%5Csqrt%202%5C%5C-%5Csqrt%202)
In fact, I can find a formula for them.
The two numbers will always be of the form:
![x_1 =a+b\sqrt c \\x_2 =a-b\sqrt c \\](https://tex.z-dn.net/?f=x_1%20%3Da%2Bb%5Csqrt%20c%20%5C%5Cx_2%20%3Da-b%5Csqrt%20c%20%5C%5C)
A and B are rational
C is rational and not a perfect square of a rational number
These are irrational because the square root will not resolve to a rational number.
12=(+√)(−√)=2−(√)2=2−2 . Looks rational to me.
1+2=+√+−√=2 , which also is rational.
Hope this helped!
Answer:
use the formula y2 - y1 /x2 - x1
y1 = 6
y2 = 9
x1 = 1
x2 = 4
Step-by-step explanation:
9 - 6/ 4 - 1
3/3
=1
slope is 1
Answer:
10 quarters, 9 dimes, and 20 nickels.
Step-by-step explanation:
q - 1 = d, 2q = n or q = 1/2n
.25q + .10(q - 1) + .05(2q) = 4.40
.25q + .10q - .10 + .10q = 4.40
.45q - .10 = 4.40
.45q = 4.50
4.5/.45 = 10
He has 10 quarters, 9 dimes, and 20 nickels.
The least number of each pack that Martin can buy is 5 packs of burger and 3 packs of bread buns.
Since the packs of bread buns contain 40 <em>bread buns </em>and each pack of burgers contains <em>24 burgers,</em> then we'll calculate the multiples of each number. This will be:
Multiples of 24 = 24, 48, 72, 96, 120
Multiples of 40 = 40, 80, 120, ...
Therefore, the least common multiple is 120.
The least number of packs of burger to buy is 5 and least number of packs of bread buns to buy is 3.
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