Answer:
Area of circle A =113.14 mm²
Area of circle b = 314.29 mm²
Area of circle C = 452.57 mm²
Area of circle A = 254.57 mm²
2.25 times
Step-by-step explanation:
Area of a circle = nr²
where n = 22/7
r = radius
Circle A's radius = 6mm
Circle B's radius = 6mm + 4mm = 10mm
Circle C's radius = 10mm + 2mm = 12mm
Circle D's radius = 12mm - 3mm = 9mm
Area of circle A = (22/7) x 6² = 113.14 mm²
Area of circle b = (22/7) x 10² = 314.29 mm²
Area of circle C = (22/7) x 12² = 452.57 mm²
Area of circle A = (22/7) x 9² = 254.57 mm²
Number of times the area of circle D is greater than that circle A = Area of circle D / Area of circle A
254.57 mm² / 113.14 mm² = 2.25 times
Answer:
y=3x-1
Step-by-step explanation:
Answer:
x <4
Step-by-step explanation:
2x-9<-1
Add 9 to each side
2x-9+9<-1+9
2x < 8
Divide each side by 2
2x/2 < 8/2
x <4
Given
s=16t^2
where
s=distance in feet travelled (downwards) since airborne with zero vertical velocity and zero air-resistance
t=time in seconds after release
Here we're given
s=144 feet
=>
s=144=16t^2
=>
t^2=144/16=9
so
t=3
Ans. after 3 seconds, the object hits the ground 144 ft. below.
Answer:
4
Step-by-step explanation:
- If you take the lowest half of the data (from the lower extreme to the median) and find the median of these numbers, that is called the lower quartile or quartile 1.
- If you take the upper half of the data (from the upper extreme to the median) and find the median of these numbers, that is called the upper quartile or quartile 3.
- The interquartile range is the difference between quartile 1 and quartile 3.
- 4, 5, 6, 8, 9, 10, 11, 13
- Median=8.5
- Q1=6.5
- Q3=10.5
- 10.5-6.5=4
I also added my PowerPoint if you didn't get what I was saying.
