5,7,8 they all equal 21, 21 divided by 3 is 7
Answer: ↓↓↓Btw sorry its not in order
Step-by-step explanation:
m∠62 and ∠g are vertical angles so they're congruent.
∠g = 62°
∠f and m∠62 are linear pairs so they're supplementary
∠f + 62 = 180
∠f = 118°
∠e is a right angle and is supplementary to a 90° angle
∠e = 90°
Adding ∠e, m∠28° and the missing angle should sum up to 180° since its a triangle. x = missing angle
x + ∠e + m∠28 = 180°
x + 90 + 28 = 180
x + 118 = 180
x = 62°
∠h ≅ x
∠h ≅ 62
∠h = 62°
∠b and m∠28 are vertical angles so they're congruent.
∠b ≅ m∠28
∠b = 28°
∠b and ∠a are linear pairs so they are supplementary.
∠a + ∠b = 180
∠a + 28 = 180
∠a = 152°
∠c and ∠a are corresponding angles so they're congruent.
∠c ≅ ∠a
∠c ≅ 152
∠c = 152°
∠c and ∠j are vertical angles so they're congruent.
∠j ≅ ∠c
∠j ≅ 152
∠j = 152°
The corresponding angle of c is 152° as well. Let's call it y. Y corresponds to the angle with a bisector. Call that whole angle z.
z ≅ y
z ≅ 152
z = 152°
Z has a bisector meaning the two angles that formed because of the bisector are congruent.
∠d and the other angle add up to 152°. They're equal.
∠d = 152 ÷ 2
∠d = 76°
Hope I helped!
Answer:
Step-by-step explanation:
Let
Y ----> field of vision that Yash's camera would need
we know that
Applying the law of sines
Solve for sin(Y)
![Y=sin^{-1}[\frac{sin(41\°)}{30}(25)]](https://tex.z-dn.net/?f=Y%3Dsin%5E%7B-1%7D%5B%5Cfrac%7Bsin%2841%5C%C2%B0%29%7D%7B30%7D%2825%29%5D)
The Acute Angles Of An Isosceles Triangle Never Add Up To 90°.
Let: U = {a,b,c,d,e,f,g,h} A = {a,c,f} B = {b,c,f} C = {b,d,e,g,h} Determine the (A∩C). A. (A∩B) ∪ (A∩C) = m B. (A∩B) ∪ (A∩C) i
DedPeter [7]
The ordering of the questions here is kind of confusing, so I'll just extract what seems to be relevant. They seem like True/False questions, but that's not clear to me.
There are no common elements between the sets <em>A</em> and <em>C</em>, so
<em>A</em> ∩ <em>C</em> = { } (the empty set)
Both sets <em>A</em> and <em>B</em> share the elements <em>c</em> and <em>f</em>, so
<em>A</em> ∩ <em>B</em> = {<em>c</em>, <em>f </em>}
and since we know <em>A</em> ∩ <em>C</em> is empty, we end up with
(<em>A</em> ∩ <em>B</em>) ∪ (<em>A</em> ∩ <em>C</em>) = {<em>c</em>, <em>f</em> } ∪ { } = {<em>c</em>, <em>f</em> }
