Answer:
Solution given:
1.
diameter(d)=6mm
base(b)=8mm
height (h)=5mm
Area of figure=area of parallelogram +area of semi circle
- base*height+½π(d/2)²
- 8*5+½*π×(6/2)²
- 40+14.14
- 54.4mm²
- <u>Area</u><u> </u><u>:</u><u>5</u><u>4</u><u>.</u><u>1</u><u>4</u><u>m</u><u>m</u><u>²</u>
2.
for triangle
base[b]=6ft
height(h)=9ft
for square
length[l]=9ft
Area of figure=area of square +area of triangle
- =l²+½*b*h
- =9²+½*6*9
- =81+27
- =108ft²
- <u>Area</u><u>:</u><u> </u><u>1</u><u>0</u><u>8</u><u>f</u><u>t</u><u>²</u>
Answer:
0.125....................
Answer:
A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds. 4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds
Step-by-step explanation:
The given table is presented as follows;
The number of laps in the range 82 to 84 seconds = 1
The number of laps in the range 84 to 86 seconds = 4
The number of laps in the range 86 to 88 seconds = 2
The number of laps in the range 88 to 90 seconds = 4
The number of laps in the range 90 to 92 seconds = 6
The number of laps in the range 92 to 94 seconds = 5
The number of laps in the range 94 to 96 seconds = 2
The number of laps in the range 96 to 98 seconds = 0
Therefore, the histogram that represents Blanca's lap times for the three days of practice is described as follows;
A graph shows lap time (seconds) labeled 82 to 98 on the horizontal axis and the number of laps on the vertical axis. 1 lap was 82 to 84 seconds. 4 laps were 84 to 86 seconds. 2 laps were 86 to 88 seconds. 4 laps were 88 to 90 seconds. 6 laps were 90 to 92 seconds. 5 laps were 92 to 94 seconds. 2 laps were 94 to 96 seconds. 0 laps were 96 to 98 seconds
Find the missing digit of the following UPC code<br>0, 2, 8, 5, 0 ,0, 1, 1, 0, 7, 0 ■
olga nikolaevna [1]
It’s the square and the number woidl be 0