Here we will tell you what 329 is rounded to the nearest hundred and also show you what rules we used to get to the answer. First, 329 rounded to the nearest hundred is:
300
<span>Remember, we did not necessarily round up or down, but to the hundred that is nearest to 329.</span>
Answer:
x = - 8 , x = 2
Step-by-step explanation:
the absolute value function always gives a positive result, however, the expression inside can be positive or negative, , that is
x + 3 = 5 ( subtract 2 from both sides )
x = 2
or
-(3 + x) = 5
- 3 - x = 5 ( add 3 to both sides )
- x = 8 ( multiply both sides by - 1 )
x = - 8
FOUND THE COMPLETE QUESTION IN ANOTHER SOURCE.ATTACHED IMAGE. For this case what we have is the following:
For the two semicircles we can model it as a complete circle.
We have to then:
Perimeter:
P = 2 * pi * r
or
P = pi * d
Where,
r = radius
d = diameter
Therefore the perimeter is:
P = 10 * pi
For the largest circle we have:
radius = 10
Perimeter:
P '= 2pi10
P '= 20pi
1/4 since 1/4 circle:
P '' = 20pi / 4 = 5pi
Then, the total perimeter of the source is:
Pt = P + P '' = 10pi + 5pi = 15pi
Pt = 15 * (3.141592)
Pt = 47.1239
round
Pt = 47.1 ft
Area:
The total area will be:
A = A (two semicircles) + A (quarter big circle)
A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
A = 98.17477042 feet ^ 2
Round:
A = 98.2 feet ^ 2
Answer:
Perimeter of the source:
Pt = 47.1 ft
Area of the source:
A = 98.2 feet ^ 2
Option C is your answer.
1 3/5 is the same as 8/5
3(8/5) = 24/5 = 4 and 4/5
The average between 12.00 and 15.00 seconds
can be defined in a few different ways.
Way #1: (speed)
Average speed =
(1/2) (the speed at 12 seconds + the speed at 15 seconds) .
Way #2: (speed)
Average speed =
(1/3) (the distance covered between 12.00 and 15.00 seconds)
Way #3: (velocity)
Average velocity =
(1/3) (distance between location at 12 sec and location at 15 sec)
in the direction from location at 12 sec to location at 15 sec.
That's the best we can do with the combination of given and
not given information in the question.