<u>Q</u><u>uest</u><u>ion</u><u>:</u>
To Simplify:
118 {121÷(11×11)-(-4)-(3-7)}
<u>Solu</u><u>tion</u>:
↠118 {121÷121+4-(-4)}
↠118 {1+4+4}
↠118 {5+4}
↠118{9}
↠118×9
↠1062
▬▬▬▬▬▬▬▬▬▬▬▬
The upper Solution is done by applying BODMAS
<u>Abou</u><u>t</u><u> </u><u>BODMAS</u><u>:</u>
B→ Bracket
O→ Of
D→ Division
M→ Multiplication
A→ Addition
S→ Subtraction
Answer:
D
Step-by-step explanation:
The range is the values of y covered by the graph
The graph on the left shows y going to - ∞
The graph has its maximum value at 4 ( turning point )
The range is therefore
- ∞ < y ≤ 4
Answer:
42 times
Step-by-step explanation:
Since there are 6 sides on a fair dice, each having a number 1 thru 6 with no numbers repeating and half those numbers being odd, Bridget can expect to roll an odd number 84*(1/2) times. So our answer is 42.
Also, if you could please mark this "brainliest" (the crown) it would help a lot!
Answer:
First pilot will gain altitude faster by 7.16 mph than second pilot.
Step-by-step explanation:
In this question we will calculate the y component of the velocity by which the pilots are gaining height. Means we will analyze which pilot will gain the altitude early.
For the first pilot vertical component of the velocity will be
v = 400 × sin30° = 400 × (1/2) = 200 mph
For second pilot vertical component of the speed is
v' = 300 × sin40° = 300 × (0.642) = 192.84 mph
Therefore pilot 1 will gain altitude faster then pilot 2 by
200 - 192.84 = 7.16 mph.
