70 ft per sec going upward 3 sec from 6 ft high should reach 216 ft, but unfortunately gravity is pulling it and at some point it starts to go backward (at 2.176 second), so to compensate the upward slowing down and distance dropped, use -32.176 times 3 square times 1/2 = -144.792 which is the natural drop distance if the football is released mid air without a initial speed after 3 seconds. 216 - 144.792 = 71.208 ft high.
Answer:
0
Step-by-step explanation:
3ab + 5b - 6 when a = -1 and b is 3
→ First substitute a and b into 3ab
3 × -1 × 3 = -9
→ Substitute b into 5b
5 × 3 = 15
→ Connect all the terms together
-9 + 15 - 6
→ Simplify
0
Let's solve this problem step by step.
28=8b+13b-35
Step 1: Bring 35 to 28.
28=8b+13b-35
+35 +35
63=8b+13b
Step 2: Add 8b and 13b.
63=21b
Step 3: Divide both sides by 21.
63/21=21b/21
So, the answer for this problem is 3=b.
Answer:
53°
Step-by-step explanation:
It is given that the total measurement of the two angles combined would equate to 116°.
It is also given that m∠WXY is 10° more then m∠ZXY.
Set the system of equation:
m∠1 + m∠2 = 116°
m∠1 = m∠2 + 10°
First, plug in "m∠2 + 10" for m∠1 in the first equation:
m∠1 + m∠2 = 116°
(m∠2 + 10) + m∠2 = 116°
Simplify. Combine like terms:
2(m∠2) + 10 = 116
Next, isolate the <em>variable</em>, m∠2. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.
First, subtract 10 from both sides of the equation:
2(m∠2) + 10 (-10) = 116 (-10)
2(m∠2) = 116 - 10
2(m∠2) = 106
Next, divide 2 from both sides of the equation:
(2(m∠2))/2 = (106)/2
m∠2 = 106/2 = 53°
53° is your answer.
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Keeping in mind that there are 5280 ft in 1 mile, and there are 60 minutes in 1 hr and 60 seconds in 1 minute, therefore 60*60 seconds in 1 hr, or 3600 seconds, then.
