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Veseljchak [2.6K]
2 years ago
14

Players from the local Little League plans to hold a car wash as a fundraiser. If they charge $5 per car, how many cars do they

need to wash to earn at least $250 dollars?
Mathematics
1 answer:
Andrews [41]2 years ago
6 0

250/5 = 50

therefore they must wash at least 50 cars

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What is mila’s average swimming speed?
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she swam 1500 meters in 1/4 of the time and he swam 750 in half of his time. Hope this helps

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Which ones have a relationship with functions ​
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b. Yes

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Identify the correct logical reason for the next step in solving this equation. 3(2 − 3x) + 4x = x − 7 Question 2 options: Multi
suter [353]

Answer:

Distributive property

Step-by-step explanation:

3(2 − 3x) + 4x = x − 7

Distribute the 3 to each term in the parentheses using the distributive property

6 -9c +4x = x-7

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3 years ago
Determine if the segment X is half the length, double the length, two thirds the length, or the same length as segment Y.
Ipatiy [6.2K]
The right answer is The Same Length.
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3 years ago
A basketball player, standing near the basket to grab a rebound, jumps 66.1 cm vertically. How much time does the player spend i
Zepler [3.9K]

Answer:

0.293 s

Step-by-step explanation:

Using equations of motion,

y = 66.1 cm = 0.661 m

v = final velocity at maximum height = 0 m/s

g = - 9.8 m/s²

t = ?

u = initial takeoff velocity from the ground = ?

First of, we calculate the initial velocity

v² = u² + 2gy

0² = u² - 2(9.8)(0.661)

u² = 12.9556

u = 3.60 m/s

Then we can calculate the two time periods at which the basketball player reaches ths height that corresponds with the top 10.5 cm of his jump.

The top 10.5 cm of his journey starts from (66.1 - 10.5) = 55.6 cm = 0.556 m

y = 0.556 m

u = 3.60 m/s

g = - 9.8 m/s²

t = ?

y = ut + (1/2)gt²

0.556 = 3.6t - 4.9t²

4.9t² - 3.6t + 0.556 = 0

Solving the quadratic equation

t = 0.514 s or 0.221 s

So, the two time periods that the basketball player reaches the height that corresponds to the top 10.5 cm of his jump are

0.221 s, on his way to maximum height and

0.514 s, on his way back down (counting t = 0 s from when the basketball player leaves the ground).

Time spent in the upper 10.5 cm of the jump = 0.514 - 0.221 = 0.293 s.

5 0
2 years ago
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