Answer:
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Step-by-step explanation:
Answer:
Point of x-intercept: (2,0).
Point of y-intercept: (0,6).
Step-by-step explanation:
1. Finding the x-intercept.
This point is where the graph of the function touches the x axis. It can be found by substituting the "y" for 0. This is how you do it:

Hence, the point of x-intercept: (2,0).
2. Finding the y-intercept.
This point is where the graph of the function touches the y axis. It can be found by substituting the "x" for 0. This is how you do it:

Hence, the point of y-intercept: (0,6).
Let us recall parallelogram properties, which states that opposite angles of parallelogram are congruent.
We can see from graph that side US is parallel to TR and measure of angle U equals to measure of angle R, therefore, quadrilateral drawn in our given graph is a parallelogram.
Since we know that opposite sides of parallelogram are congruent. In our parallelogram UT=SR and US=TR.
In our triangle STU and triangle TSR side TS=TS by reflexive property of congruence.
Therefore, our triangles are congruent by SSS congruence.
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Answer:
y -2 = 2(x -2)
Step-by-step explanation:
The slope of the line is given by the slope formula:
m = (y2 -y1)/(x2 -x1)
m = (4 -2)/(3 -2) = 2/1 = 2
The point-slope form of the equation for the line is ...
y -k = m(x -h) . . . . . . line with slope m through point (h, k)
For the first point and the slope we found, the equation is ...
y -2 = 2(x -2)
__
You can rearrange this to any form you may like.
y -2 = 2x -4 . . . eliminate parentheses
y = 2x -2 . . . . . slope-intercept form
2x -y = 2 . . . . . standard form
Answer:
The expression
represents the number
rewritten in a+bi form.
Step-by-step explanation:
The value of
is
in term of ![i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}](https://tex.z-dn.net/?f=i%5E%7B2%7D%5B%5Ctex%5D%20can%20be%20written%20as%2C%20%3C%2Fp%3E%3Cp%3E%5Btex%5Di%5E%7B4%7D%3Di%5E%7B2%7D%5Ctimes%20i%5E%7B2%7D)
Substituting the value,

Product of two negative numbers is always positive.

Now
in term of ![i^{2}[\tex] can be written as, [tex]i^{3}=i^{2}\times i](https://tex.z-dn.net/?f=i%5E%7B2%7D%5B%5Ctex%5D%20can%20be%20written%20as%2C%20%3C%2Fp%3E%3Cp%3E%5Btex%5Di%5E%7B3%7D%3Di%5E%7B2%7D%5Ctimes%20i)
Substituting the value,

Product of one negative and one positive numbers is always negative.

Now
can be written as follows,

Applying radical multiplication rule,


Now,
and 

Now substituting the above values in given expression,

Simplifying,

Collecting similar terms,

Combining similar terms,

The above expression is in the form of a+bi which is the required expression.
Hence, option number 4 is correct.