Answer:
y- intercept --> Location on graph where input is zero
f(x) < 0 --> Intervals of the domain where the graph is below the x-axis
x- intercept --> Location on graph where output is zero
f(x) > 0 --> Intervals of the domain where the graph is above the x-axis
Step-by-step explanation:
Y-intercept: The y-intercept is equivalent to the point where x= 0. 'x' is the input variable in an equation, therefore the y-intercept is where the input, or x, is equal to 0.
f(x) <0: Notice the 'lesser than' sign. This means that the value of f(x), or 'y', is less than 0. This means that this area consists of intervals of the domain below the x-axis.
X-intercept: The x-intercept is the location of the graph where y= 0, or the output is equal to 0.
f(x) >0: In this, there is a 'greater than' sign. This means that f(x), or 'y', is greater than 0. Therefore, this consists of intervals of the domain above the x-axis.
Answer:19+3x
Step-by-step explanation:
4+3(5+x)
Open bracket
4+15+3x
19+3x
Not unless the digits in each period are also in the same order. That is, the digit in each PLACE must be the same.
Answer:
The values of x and y in the diagonals of the parallelogram are x=0 and y=5
Step-by-step explanation:
Given that ABCD is a parallelogram
And segment AC=4x+10
From the figure we have the diagonals AC=3x+y and BD=2x+y
By the property of parallelogram the diagonals are congruent
∴ we can equate the diagonals AC=BD
That is 3x+y=2x+y
3x+y-(2x+y)=2x+y-(2x+y)
3x+y-2x-y=2x+y-2x-y
x+0=0 ( by adding the like terms )
∴ x=0
Given that segment AC=4x+10
Substitute x=0 we have AC=4(0)+10
=0+10
=10
∴ AC=10
Now (3x+y)+(2x+y)=10
5x+2y=10
Substitute x=0, 5(0)+2y=10
2y=10
∴ y=5
∴ the values of x and y are x=0 and y=5
Answer:
y we use y
Step-by-step explanation:
