∠1 and ∠2 form a straight angle. ∠1 has a measure of 28°. What is the measure of ∠2? PLZ HELP

Mathematics

2 answers:

Natalka [10]2 years ago

6 0

Answer:

∠2 = 152

Step-by-step explanation:

A straight angle should always equal 180, and that is really the key to how we answer this problem. That means that ∠1 and ∠2 are supplementary, and add up to 180°.

That means that 180-28, will give us our other angle.

Zolol [24]2 years ago

5 0

Answer:

152°

Step-by-step explanation:

Angles on a straight line add up to 180°

Angle 2 + 28 = 180

Angle 2 = 180 - 28

= 152°

You might be interested in

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.