Compare and contrast both densely-ionizing and sparsely-ionizing radiation.
1 answer:
You might be interested in
<u>For the first question we use point-slope form of an equation</u>
y-
=m(x-
)
y=m(x-)+
then we plug in the known values
y=m(x-4)+7
we leave the slope as variable m since it is undefined
<u>For Question 2
</u>
A line that is parallel to the x-axis is a horizontal line and since the slope of a line is defined as the Δy/Δx (change in y/change in x) and the change in y is 0 at any 2 points observed on the line the slope is 0. (0/any number is 0)
<u>For Question 3
</u>
Following the same relationship as question 2 we can solve for the slope.
Δy/Δx
-
/-
now we plug in the known values from the two points given
(5-7)/(-3-1)
-2/-4
m=1/2 or 0.5
<u><em>and For the Final Question</em></u>
a translation left or right is done by affecting the x variable if you add 2 to x then the x value will have to be 2 less to get the same result...in other words when x is 1 the value of y is also 1...but if I wanted the whole equation translated left 2 unites then I would want the same y-value at an x-value 2 smaller... in other words, in our example x will be -1 when y is 1. For this value found on the graph to match the equation our x value must have 2 added to it in the equation....therefore the equation that translates y=|x| two units left is...
y=|x+2|
y-
y=m(x-
then we plug in the known values
y=m(x-4)+7
we leave the slope as variable m since it is undefined
<u>For Question 2
</u>
A line that is parallel to the x-axis is a horizontal line and since the slope of a line is defined as the Δy/Δx (change in y/change in x) and the change in y is 0 at any 2 points observed on the line the slope is 0. (0/any number is 0)
<u>For Question 3
</u>
Following the same relationship as question 2 we can solve for the slope.
Δy/Δx
now we plug in the known values from the two points given
(5-7)/(-3-1)
-2/-4
m=1/2 or 0.5
<u><em>and For the Final Question</em></u>
a translation left or right is done by affecting the x variable if you add 2 to x then the x value will have to be 2 less to get the same result...in other words when x is 1 the value of y is also 1...but if I wanted the whole equation translated left 2 unites then I would want the same y-value at an x-value 2 smaller... in other words, in our example x will be -1 when y is 1. For this value found on the graph to match the equation our x value must have 2 added to it in the equation....therefore the equation that translates y=|x| two units left is...
y=|x+2|