Answer:
1) ∠A=84°
2) ∠C=20°
Step-by-step explanation:
1)
First, find ∠C:
<em>(I'm assuming the exterior angle of 126° makes a straight line with ∠C)</em>
The angles on a straight line always add up to 180. Therefore:
∠C+126=180
∠C=180-126
∠C=54
Then find ∠B:
We also know that all the angles in a triangle add up to 180. Therefore:
∠A+∠B+∠C=180
∠A+∠B+54=180
∠A+∠B=126
<em>(we know ∠A=2(∠B))</em>
2(∠B)+∠B=126
3(∠B)=126
∠B=42
Now, find ∠A:
∠A=2(∠B)
∠A=2(42)
∠A=84°
2)
First, find ∠B:
<em>(Again, I'm assuming the exterior angle of 100° makes a straight line with ∠B)</em>
The angles on a straight line always add up to 180. Therefore:
∠B+100=180
∠B=180-100
∠B=80
Then find ∠A:
We also know that all the angles in a triangle add up to 180. Therefore:
∠A+∠B+∠C=180
∠A+80+∠C=180
∠A+∠C=100
<em>(we know ∠A=4(∠C))</em>
4(∠C)+∠C=100
5(∠C)=100
∠C=20°
Step-by-step explanation:
Answer:
Step-by-step explanation:
l^2=[(7-V2)/2]^2+(9-V2/2)^2=
49-7V2+2/4+81-9V2+2/4=
49+81+4/4-16V2=
131-16V2=108.4 ft
so l=V108.4≈10.4ft
Answer:
P43=4!(4–3)!=241=24
Step-by-step explanation:
There are four choices you can make for the lead reindeer. For each possible choice, there are then three remaining you can choose to fly second, making 4×3=12 choices for the lead pair. For each possible choice there are two remaining reindeer to take up the back position, making 12×2=24 choices for the team of three.
This type of problem is called a permutation problem, and we write Pnr for the number of ways of choosing r items from n possibilities when the order of the items matters. In this case we are choosing 3 reindeer from 4 possibilities, and the order they appear in the flying line does matter, so the answer we want is P43. The general formula is Pnr=n!(n−r)!. For the answer we are looking for we therefore have:
P43=4!(4–3)!=241=24
Answer:
5
Step-by-step explanation:
-1+(1)= 0
0+(4)=4
1+4=5
