Helppp!!!! please!!!

Mathematics

2 answers:

Ulleksa [173]3 years ago

5 0

Answer:

$\boxed{Area = 32.5 mm^2}$

Step-by-step explanation:

Area of Triangle = $\frac{1}{2} (Base)(Height)$

Where base = 13 mm, Height = 5 mm

Area = 1/2 (13)(5)

=> Area = 1/2 (65)

=> Area = 32.5 mm^2

vodka [1.7K]3 years ago

3 0

Answer:

c. 32.5 mm²

Step-by-step explanation:

The equation for the area of a triangle is a=1/2bh. In this equation you basically multiply the length of the base of the triangle times the height of the triangle and divide that by 2 to get the area. So, you would take 13, the base, and multiply it by 5, the height to get 65. Then, you would divide that by 2 to get 32.5 mm².

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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.