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:
- 12 ft parallel to the river
- 6 ft perpendicular to the river
Step-by-step explanation:
The least fence is used when half the total fence is parallel to the river. That is, the shape of the rectangle is twice as long as it is wide.
72 = W(2W)
36 = W²
6 = W . . . . . . the width perpendicular to the river
12 = 2W . . . . the length parallel to the river
_____
<em>Development of this relation</em>
Let T represent the total length of the fence for some area A. Then if x is the length along the river, the width is y=(T-x)/2, and the area is ...
A = xy = x(T -x)/2
Note that the equation for area is that of a parabola with zeros at x=0 and at x=T. That is, for some fence length T, the area will be a maximum at the vertex of this parabola. That vertex is located halfway between the zeros, at ...
x = (0 +T)/2 = T/2
The corresponding area width (y) is ...
y = (T -T/2)/2 = T/4
Equivalently, the fence length T will be a minimum for some area A when x=T/2 and y=T/4. This is the result we used above.
When you have something like this, all you need to do is substitute the values, the last is for what value of x
For the first one;
((x^2+1)+(x-2))(2)
(x^2+x-1)(2)
(2)^2+(2)-1
4+2-1
5
For the second one;
((x^2+1)-(x-2))(3)
(x^2-x+3)(3)
(3)^2-(3)+3
9-3+3
9
For the last one;
3(x^2+1)(7)+2(x-2)(3)
3((7)^2+7)+2((3)-2)
3(49+7)+2(3-2)
3(56)+2(1)
168+2
170
Answer:
46.7%
Step-by-step explanation:
Given:
Total number of students in Mrs. Verner's class = 15
Number of girls = 8
To find: percentage are boys
Solution:
Percentage of boys = ( Number of boys / Total number of students ) × 100
Number of boys = Total number of students - Number of girls = 15 - 8 = 7
So,
Percentage of boys = × 100 = 46.7%
