Step-by-step explanation:
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Write the equation of the line that passes through the points (0, 3) and (2. 6)
1 answer:
Answer:
Y=3/2X + 3
Step-by-step explanation:
Y= mX + b
b is the y intercept or the Y number for when the point is in place o for X
(0, 3) is the Y intersect so b is 3
this is shown on the diagram I drew below
m is the slope of the line and can be calculated
m = change in Y/ Change in X
Our change in Y is 6-3
And out change is X is 2-0
So
m = 3/2
This is shown on the image below
Y=3/2X + 3
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1. From your description, I can infer that the multiplication is:
The first thing we are going to do is simplify the radicands 196 ans 108 (picture 1):
and 
Knowing this, we can rewrite our radicals as follows:
Remember that![\sqrt[n]{x^n} =x](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5En%7D%20%3Dx)
; in other words if the radicand is raised to the same power as the index of the radical, we can take the radicand out. Since 2 and 7 are raised to the power 2 and the index of the radical is also 2 (square root), we can take out 2 and 7:
Look! we have the same numerator and denominator in our fraction, so we can cancel them both:
Notice that we can write
as 
, so we can rewrite our expression one last time:
We can conclude that the correct option is:
2. The <span>product of a nonzero rational number and an irrational number is always an irrational number.
Proof by contradiction:
Lets assume that the product of an irrational number and a rational non-zero number is always rational.
Let </span>
be and irrational number and let 
and 
be two rational numbers with 
, 
, 
, and 
are non-zero integers.
Since integers are closed under multiplication,
is a rational number. Since is an irrational number and , we have a logical contradiction, so we can conclude that the product of an irrational number and a rational non-zero number is always an irrational number.
The first thing we are going to do is simplify the radicands 196 ans 108 (picture 1):
Knowing this, we can rewrite our radicals as follows:
Remember that
Look! we have the same numerator and denominator in our fraction, so we can cancel them both:
Notice that we can write
We can conclude that the correct option is:
2. The <span>product of a nonzero rational number and an irrational number is always an irrational number.
Proof by contradiction:
Lets assume that the product of an irrational number and a rational non-zero number is always rational.
Let </span>
Since integers are closed under multiplication,