For this case, the first thing we must do is define variables.
We have then:
t: the time in minutes
k: the number of kilometers
The relationship between both variables is direct.
Therefore, the function is:
Where, "c" is a constant of proportionality.
To determine "c" we use the following data:
After running for 18 minutes, she completes 2 kilometers.
Substituting values:
Clearing c we have:
Then, the equation is given by:
Answer:
An equation that can be used to represent k, the number of kilometers Julissa runs in t minutes is:
. . . . . . .. . . . . . . . . . . . . . . . . . . . ./ . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer:
The graph that includes points (-3,-3) and (0,3)
Step-by-step explanation:
In the pictures attached, the options are shown.
The equation:
y+3=2(x+3)
has the point-slope form, which is:
y-y₁=m(x-x₁)
where (x₁, y₁) is a point on the line and <em>m</em> is its slope. This means that (-3,-3) is on the line. To know the y-intercept of the line, we have to replace x = 0 into the equation, as follows:
y+3=2(0+3)
y+3 = 6
y = 6 - 3
y = 3
Then, point (0, 3) is on the line.
Answer:
The graph is sketched by considering the integral. The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.
Step-by-step explanation:
We sketch the integral ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ. We consider the inner integral which ranges from r = 0 to r = 6/cosθ. r = 0 is located at the origin and r = 6/cosθ is located on the line x = 6 (since x = rcosθ here x= 6)extends radially outward from the origin. The outer integral ranges from θ = 0 to θ = π/4. This is a line from the origin that intersects the line x = 6 ( r = 6/cosθ) at y = 1 when θ = π/2 . The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.
