In equation 1, When x = 0, y = -3 ( 0, -3)
And when y = 0, x = -3....( 3,0)
In equation 2, when x = 1, y = -4(1,-4)
and when y = 2, x = 5....(5,2)
You have to plot the points on graph paper and the solutions are where the two lines intersect.
You can use any points for x and y.
Answer:
Step-by-step explanation:
y intercept please scroll down--i need to see the rest of the graph-- y intercept is where the graph intersects with the y axis
x intercepts are x=2 and x=6
greatest value of y is y=2 and it occurs when x=4
for x between... what are the options?
if the options are greater than, equal to, or less than, the answer is greater than
Yes it can just keep dividing by 3 until you can't anymore
Solution,
Given; x^3-7x^2-x+7=0
Using factor theorem,
we have,
i)x^3-7x^2-x+7=0
let's check whether 1 is the factor of given equation or not by replacing the value of x by one..
or, 1^3-7.1^2-1+7=0
or, 1-7-1+7=0(.°. Cancellation of +1&-1 and +7&-7 in equation)
.°.0=0
which proves that 1 is one of the factor of x
therefore(x-1) is a factor of x^3-7x^2-x+7=0.
Now,
Expanding the x^3-7x^2-x+7=0 to get (x-1) as a factor...
ii)x^3-7x^2-x+7=0
or,x^3-x^2-6x^2+6x-7x+7
or,x^2(x-1) - 6x(x-1) - 7(x-1) = 0
or,(x-1) (x^2-6x-7)w = 0
or,,(x-1) (x^2-(7-1)x-7) =0
or,,(x-1) (x^2-7x + x -7) =0
or,(x-1) [x(x-7)+1(x-7)] = 0
or,(x-1) (x+1) (x-7) = 0
Either,
i) x - 1=0
.°.x = 1
And,
ii)x + 1 = 0
.°. x = -1
Or,
iii) x - 7 = 0
.°.x =7
Hence, the required value of x are -1 ,+1 & 7.
Step-by-step explanation:
Open the brackets ,
=> 11( x + 2 )
=> 11x + 11*2
=> 11x + 22
