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hodyreva [135]
3 years ago
5

The weight of the rectangle is 80

Mathematics
1 answer:
Helga [31]3 years ago
7 0

Answer:

6 2/3

Step-by-step explanation:

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Answer:

10 units

Step-by-step explanation:

The distance from the point (0,0) to (10,0) is 10 units

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-2

Step-by-step explanation:

Two points (1,0) (0,2) use slope formula 0-2/1-0

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A certificate of deposit has an annual simple interest rate of 5.25%. If $567 in interest is earned over a 6 year period, how mu
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Find the the amount of interest per year. Since $567 were earned over 6 years, you divide the interest earned by the number of years it took to accumulate it:

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3 years ago
A circle is centered at J(3, 3) and has a radius of 12.
stealth61 [152]

Answer:

(-6,\, -5) is outside the circle of radius of 12 centered at (3,\, 3).

Step-by-step explanation:

Let J and r denote the center and the radius of this circle, respectively. Let F be a point in the plane.

Let d(J,\, F) denote the Euclidean distance between point J and point F.

In other words, if J is at (x_j,\, y_j) while F is at (x_f,\, y_f), then \displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}.

Point F would be inside this circle if d(J,\, F) < r. (In other words, the distance between F\! and the center of this circle is smaller than the radius of this circle.)

Point F would be on this circle if d(J,\, F) = r. (In other words, the distance between F\! and the center of this circle is exactly equal to the radius of this circle.)

Point F would be outside this circle if d(J,\, F) > r. (In other words, the distance between F\! and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between J and F:

\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145}  \end{aligned}.

On the other hand, notice that the radius of this circle, r = 12 = \sqrt{144}, is smaller than d(J,\, F). Therefore, point F would be outside this circle.

5 0
3 years ago
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