Depends. All multiples of 400, you could have many different options. For example, 400 buses 1 kid on ea. 40 buses 10 kids on ea.
400x1=400
40x10=400
Just if the two numbers multiple to 400 it can be an answer
68= 4x-2+30 (notice they are vertical angles)
68= 4x+28
40=4x
10=x
68+y= 180 (forms a straight angle)
y=112
Final answer: x=10, y=112
Answer:
One-half times 10 times 24
Step-by-step explanation:
![area \: of \: triangle = \frac{1}{2} \times height \times base \\ \\ area \: of \: triangle = \frac{1}{2} \times 10 \times 24](https://tex.z-dn.net/?f=area%20%5C%3A%20of%20%5C%3A%20triangle%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20height%20%5Ctimes%20base%20%5C%5C%20%20%5C%5C%20area%20%5C%3A%20of%20%5C%3A%20triangle%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%20%20%2010%20%5Ctimes%2024)
Answer: 86 degree
Step-by-step explanation:
The Straight line is 180 degrees so if we minus 180 by the outward degree there will be the degree for inward degree
P: 95 degrees T: 160 degree S: 88 degrees R: 91 degree
The degree for the pentagon is 540 degrees so If I minus 540 by all the four inward degrees there will be the missing last degree
95 + 160 + 88 + 91 = 434 degree
540 - 434 = 106 degree
180 - 106 = 94 degree
So the last inward degree is 94
Finally, if I minus 180 by 94 (the way I used to find inward degree and to find outward degree for now) it equals 86 degree
so X is 86 degree
Sorry if this was too confusing:)
Scalene = yes it is possible (example: 105 degrees, 5 degrees, 70 degrees)
Right = No, the reason why is because the remaining angles must add to 90 which is less than a right angle. So it's impossible to have one angle over 90 degrees.
Acute = Also no. All three angles are less than 90 degrees
Isosceles = Yes. It is possible. One example would be 120 degrees, 30 degrees, 30 degrees.