Answer:
yes
Step-by-step explanation:
Answer:
1 ≥ t ≤ 3
Step-by-step explanation:
Given
h(t) = -16t² + 64t + 4
Required
Determine the interval which the bar is at a height greater than or equal to 52ft
This implies that
h(t) ≥ 52
Substitute -16t² + 64t + 4 for h(t)
-16t² + 64t + 4 ≥ 52
Collect like terms
-16t² + 64t + 4 - 52 ≥ 0
-16t² + 64t - 48 ≥ 0
Divide through by 16
-t² + 4t - 3 ≥ 0
Multiply through by -1
t² - 4t + 3 ≤ 0
t² - 3t - t + 3 ≤ 0
t(t - 3) -1(t - 3) ≤ 0
(t - 1)(t - 3) ≤ 0
t - 1 ≤ 0 or t - 3 ≤ 0
t ≤ 1 or t ≤ 3
Rewrite as:
1 ≥ t or t ≤ 3
Combine inequality
1 ≥ t ≤ 3
Right off the bat we know that the y-intercept is going to be 5 because of the second co-ordinate.
Now for the slope, because the y co-ordinate doesn't change, it has no slope so the equation is simply: y=5
Hope this helps :)
Answer:
a) there is s such that <u>r>s</u> and s is <u>positive</u>
b) For any <u>r>0</u> , <u>there exists s>0</u> such that s<r
Step-by-step explanation:
a) We are given a positive real number r. We need to wite that there is a positive real number that is smaller. Call that number s. Then r>s (this is equivalent to s<r, s is smaller than r) and s is positive (or s>0 if you prefer). We fill in the blanks using the bold words.
b) The last part claims that s<r, that is, s is smaller than r. We know that this must happen for all posirive real numbers r, that is, for any r>0, there is some positive s such that s<r. In other words, there exists s>0 such that s<r.
Answer:
The standard form of the equation of the circle is 
.
Step-by-step explanation:
A circle is the set of points in a plane that lie a fixed distance, called the radius, from any point, called the center.
The equation of a circle in standard form is
where <em>r</em> is the radius of the circle, and <em>h</em>, <em>k</em> are the coordinates of its center.
When the center of the circle coincides with the origin 
, so
We are also told that the circle contains the point (0, 1), so we will use that information to find the radius <em>r</em>.
Therefore, the standard form of the equation of the circle is .
