Answer:It's called the "y intercept" and it's the y value of the point where the line intersects the y- axis. For this line, the y-intercept is "negative 1." You can find the y-intercept by looking at the graph and seeing which point crosses the y axis. This point will always have an x coordinate of zero.
Step-by-step explanation:
The Y intercept of a straight line is simply where the line crosses the Y axis.
Example
Y intercept
In the above diagram the line crosses the Y axis at 1.
The Y intercept is equal to 1 and the point is written as (0,1). Notice that for the y-intercept the x-coordinate of the point is always zero..
Answer:
(2x-1)(x-7)
Step-by-step explanation:
Answer:
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Step-by-step explanation:
The factors of 42 : ±1 , ±2 , ±3 , ±6, ±7, ±14, ±21 , ±42
<u>Using trial and error method we will find the first root of the polynomial.</u>
1 : ( 1 )³ - 10 ( 1 )² - 53 ( 1 ) - 42
1 - 10 - 53 - 42 ≠ 0
Therefore 1 is not a root
-1 : ( -1 )³ - 10 (- 1 )² - 53 (- 1 ) - 42
- 1 - 10 + 53 - 42
53 - 53 = 0
Therefore - 1 is a root of the polynomial.
Therefore ( x + 1) is a factor.
<u>Now by long division or by using synthetic division we can find other factors.</u>
<u>Synthetic Division :</u>
-1 | 1 -10 -53 - 42
| 0 - 1 11 42
|______________________
1 - 11 - 42 0
Therefore ,
------ ( 1 )
<u>Next further factorize x² - 11x - 42</u>
x² - 11x - 42
= x² - 14x + 3x - 42
=x(x - 14) + 3 (x - 14)
=(x + 3 )(x - 14)
Therefore ,
Answer:
-3
Step-by-step explanation:
Hope this helps:)
<em>Given - a+b+c = 0</em>
<em>To prove that- </em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
<em>Now we know that</em>
<em>when x+y+z = 0,</em>
<em>then x³+y³+z³ = 3xyz</em>
<em>that means</em>
<em> (x³+y³+z³)/xyz = 3 ---- eq 1)</em>
<em>Lets solve for LHS</em>
<em>LHS = a²/bc + b²/ac + c²/ab</em>
<em>we can write it as LHS = a³/abc + b³/abc + c</em><em>³</em><em>/abc</em>
<em>by multiplying missing denominators,</em>
<em>now take common abc from denominator and you'll get,</em>
<em>LHS = (a³+b³+c³)/abc --- eq (2)</em>
<em>Comparing one and two we can say that</em>
<em>(a³+b³+c³)/abc = 3</em>
<em>Hence proved,</em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
