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Tatiana [17]
3 years ago
12

2 liters is equivalent to how many ounces

Mathematics
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

hey hon! 2 liters is equal to 67.628 fluid ounces :) hope you have a nice day.

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A basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modele
aev [14]

THIS IS THE COMPLETE QUESTION

basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modeled by the equation h = 6 + 30t – 16t2. Two-tenths of a second after the shot is launched, an opposing player leaps up to block the shot. The height of the shot blocker’s outstretched hands t seconds after he leaps is modeled by the equation h = 9 + 25t – 16t2. If the ball reaches the net 1.7 seconds after the shooter launches it, does the leaping player block the shot?

Yes, exactly 0.6 seconds after the shot is launched.

Yes, between 0.64 seconds and 0.65 seconds after the shot is launched

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

No, the shot is not blocked.

Answer:

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

Given that the

equation for shot block height h = 9 + 25t – 16t2.

equation for ball height h = 6 + 30t – 16t2.

From the question ,it can be deducted that the shot is made before two tenths of a second or 0.2seconds

CHECK THE ATTACHMENT TO COMPLETE THE DETAILED SOLUTION

5 0
3 years ago
Read 2 more answers
Need this answered ASAP
Fudgin [204]

i think is 3rd one, cuz in the app photomath it said (-1, -9)


so this might be close

5 0
3 years ago
What is 1,896,345 rounded
vazorg [7]
It is rounded to the number 2,000,000.
6 0
4 years ago
Solve:
never [62]
C. Eight ninths because 1/3=3/9 and 3/9 +5/9= 8/9
5 0
3 years ago
The hawaiian alphabet has 12 letters. How many permutations are possible for five of these letters
guajiro [1.7K]

Answer: The answer is 50980

Step by Step Explanation:

First, temporarily assume that two letters with I are different, call them i 1 and i 2. Three "a" are also called as a 1, a 2 and a 3, and two h as h 1 and h 2. Then there are 11 * 10 * 9 * 8 * 7 = 55440 possible "words" (one of 11 is the first letter, 10 is the second, and so on). But because equal letters do the same "words," some "words" were counted twice or more. We have to deduct the number of "parasitic" counts although it is fairly small. The words that counted more than once are divided into many disjoint sets: 1) with two I but without repetitions of a and h; 2) with two h but without repetitions of a and I 3) with two a but without repetitions of I and h; 4) with three a but without repetitions of I and h; 5) with two I and two a's; 6) two i's and tree a's; 7) two i's and two h's 8) two h's and two a's; 9) two h's and one tree. The first category includes terms counted twice and its scale is (5 * 4) * (6 * 5 * 4) = 2400 (the first I stays at one of the 5 positions, the second at one of the 4, then 11-2i-1h-2a=6). So we have 2400/2 = 1200 to subtract. Group 2 gives -600 as well, and group 3 also. Group 4 gives * (6 * 5) = 1800 (5 * 4 * 3), and the terms are counted 6 times, -300. Groups 5, 7 , 8: 5 * 4 * 3 * 2 * 6 = 720 and counted four times, therefore -180. Group 6 and 9: 5 * 4 * 3 * 2 * 1 = 120, with 12 counts, -10. Altogether -(1200 * 3 + 300 + 180 * 3 + 10 * 2) = -4460.The answer will be 55440-4460 = 50980.

3 0
4 years ago
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