Alright, lets get started.
If the polygon measures 45 degree exterior angle and we are asked to find number of sides, there is a formula for that.
= value of exterior angle
Where n is number of sides of ploygon

Multiplying n into both sides


Now divide 45 in both sides

n = 8
Means number of sides of polygon is 8, hence its Octagon. Answer
Now we have given 45 exterior angle hence interior angle would be
180 - 45 = 135°
So, sum of all interior angles would be = number of sides * interior angle
sum of all interior angles would be = 
Hence sum of all interior angles would be 1080°. Answer
Hope it will help :)
Answer:
I think it's true
Step-by-step explanation:
Answer:
El área de un rectángulo de largo L y ancho A es: L*A.
El área de un triangulo de alto H y base B es: H*B/2.
Ahora veamos las tres figuras:
Rectángulo amarillo:
Largo = 40m
Ancho = 45m
Área = 40m*45m = 1800 m^2
Rectángulo blanco:
Largo = 40m
Ancho = 35m
Área = 40m*35m = 1400m^2
Triangulo azul:
Base = 35m
Alto = 50m
Área = 35m*50m/2 = 875m^2
Area total = 1800 m^2 + 1400m^2 + 875m^2 = 4075m^2
Answer: b 5 calories per minute
Step-by-step explanation: you can see that the graph starts at 0 and at the one at the bottom it goes to 5 and then 10 for 2 minutes and so on
Answer:
10.5 hours.
Step-by-step explanation:
Please consider the complete question.
Working together, two pumps can drain a certain pool in 6 hours. If it takes the older pump 14 hours to drain the pool by itself, how long will it take the newer pump to drain the pool on its own?
Let t represent time taken by newer pump in hours to drain the pool on its own.
So part of pool drained by newer pump in one hour would be
.
We have been given that it takes the older pump 14 hours to drain the pool by itself, so part of pool drained by older pump in one hour would be
.
Part of pool drained by both pumps working together in one hour would be
.
Now, we will equate the sum of part of pool emptied by both pumps with
and solve for t as:








Therefore, it will take 10.5 hours for the newer pump to drain the pool on its own.