Which polynomial identity will prove that 117 = 125 − 8?
2 answers:
Answer:
Difference of cubes.
Step-by-step explanation:
We are asked to find the identify that will prove that .
Polynomial identities are equations that are true and have general applications in our algebra work.
We can see that 125 equals and 8 equals
. So polynomial identity used in our given problem is difference of cubes.
Difference of cubes:
Let us prove that 117=125-8.
We can write right hand side of our given equation as:
We can see that left hand side of our equation is equal to right hand side of equation. Hence, proved that difference of cubes will prove our equation equal.
You might be interested in
Hello!
<u>Number 22</u>: We'd plot the first point at 0 since there is no stated y-intercept. Next, we'd use our slope to determine where to plot the next point, and that would create our line. According to the problem, our slope is
, which automatically tells us that the slope would be going downwards because it's negative.
To plot our point, use the slope while going down and across from our y-intercept, which is 0. Go down 1, and over 2.
Your points should be at (0, 0) and (-1, 2)
<u /><u>Number 23:</u> This one will be a bit trickier since the equation is not in slope-intercept form. First, let's convert it to slope-intercept form.
Flip some of those numbers around to get our equation in slope-intercept form:
Now to graph this, we do the same as we did for the last problem. Plot our first point at (0, 2), since 2 is our y-intercept. Afterwards, go up 2 and over 3, then plot the other point.
Your points should be at (0, 2) and (4, 3)
<u>Number 22</u>: We'd plot the first point at 0 since there is no stated y-intercept. Next, we'd use our slope to determine where to plot the next point, and that would create our line. According to the problem, our slope is
To plot our point, use the slope while going down and across from our y-intercept, which is 0. Go down 1, and over 2.
Your points should be at (0, 0) and (-1, 2)
<u /><u>Number 23:</u> This one will be a bit trickier since the equation is not in slope-intercept form. First, let's convert it to slope-intercept form.
Flip some of those numbers around to get our equation in slope-intercept form:
Now to graph this, we do the same as we did for the last problem. Plot our first point at (0, 2), since 2 is our y-intercept. Afterwards, go up 2 and over 3, then plot the other point.
Your points should be at (0, 2) and (4, 3)