Hello! The x-intercept is where the line crosses the x-axis when y = 0. We already have the expression shown. We will plug in 0 as y to get this:
5x - 8(0) = 40.
As you probably know, anything times 0 is 0, so -8 * 0 is 0. All we have is 5x = 40. Divide each side by 5 to isolate the x. 5x/5 cancels out. 40/5 is 8. There. x = 8. The x-intercept is (8, 0).
$3.50 being the amount the item <em>t</em> costs and <em>t</em> being the number of this item.
Answer:
Domain {x : x > 1}
Range {y : y ∈ R}
Vertical asymptote x = 0
x-intercept (1, 0)
End behavior consistent
Graph attached down
Step-by-step explanation:
Let us study the equation:
∵ y = log(x)
→ It is a logarithmic function, so no negative values for x
∴ Its domain is {x : x > 1}
∴ Its range is {y : y ∈ R}, where R is the set of the real numbers
→ An asymptote is a line that a curve approaches, but never touches
∵ x can not be zero
∴ It has a vertical asymptote whose equation is x = 0
→ x-intercept means values of x at y = 0, y-intercept means
values of y at x = 0
∵ x can not be zero
∴ There is no y-intercept
∵ y can be zero
∴ The x-intercept is (1, 0)
→ The end behavior of the parent function is consistent.
As x approaches infinity, the y-values slowly get larger,
approaching infinity
∵ y = log(x) is a parent function
∴ The end behavior is consistent
→ The graph is attached down
Given:
The length of the ladder = 12 ft
The angle of ladder with ground = 60 degrees
To find:
How far up the building the ladder will reach.
Solution:
Using the given information draw a figure as shown below.
We need to find the vertical distance between the top of ladder and the ground.
Let x be the required distance.
In a right angle triangle,

In the below triangle ABC,



Multiply both sides by 12.


Therefore, the ladder will reach
ft far up the building.