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Rufina [12.5K]
3 years ago
7

4. A bag of rice at the warehouse store weighs

Mathematics
1 answer:
slega [8]3 years ago
6 0
Total number of pounds=15x number of bags
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Add -4 + 2/3<br><br>A) -5 2/3<br><br>B) -4 2/3<br><br>C) -3 2/3<br><br>D) -3 1/3
kodGreya [7K]

Answer:

D

Step-by-step explanation:

4 0
3 years ago
If Cyrus knows that △ABC∼△EDC and AB¯¯¯¯¯¯¯¯∥ED¯¯¯¯¯¯¯¯, how can he prove that line m passing through AB¯¯¯¯¯¯¯¯ has the same sl
tester [92]

As AB and ED are parallel, they have the same slope. This means that since lines that are colllinear with parallel segments are parallel, lines l and m are parallel.

5 0
2 years ago
marcos has a coin collection. His collection started with 25 coins. He adds 8 coins to his collection every month. Based on this
avanturin [10]

Answer: G does not show the relationship

Step-by-step explanation:

x is the number of months and he originally had 25 coins. The correct representation of this equation would be y=8x+25 since he adds 8 coins every month. So, G does not show the relationship.

8 0
3 years ago
For the problem, use the discriminant to determine the number of real solutions for the equation. Then, find the solutions and c
Ksivusya [100]

Answer:

Discriminant = 55.2 > 0 -> 2 real solutions

Solutions: t1 = -1.1663 s and t2 = 0.35 s

The solution t1 doesn't make sense for this problem, as we can't have a negative value for the time.

So the solution is t2 = 0.35 s

Step-by-step explanation:

To find the time when the ball will reach the height of 2 meters, we just need to use the value of h = 2 in the equation given. So, we have that:

−4.9t^2 − 4t + 4 = 2

−4.9t^2 − 4t + 2 = 0

For this equation, we have the constants a = -4.9, b = -4 and c = 2. So the discriminant Delta is:

Delta = b^2 - 4ac = 16 + 39.2 = 55.2

sqrt(Delta) = 7.4297

As Delta > 0, we have 2 real solutions

t1 = (-b + sqrt(Delta)) / 2a = (4 + 7.4297) / (-9.8)  = -1.1663 s

t2 = (-b - sqrt(Delta)) / 2a = (4 - 7.4297) / (-9.8)  = 0.35 s

Number of real solutions: 2

Solutions: t1 = -1.1663 s and t2 = 0.35 s

The solution t1 doesn't make sense for this problem, as we can't have a negative value for the time.

So the solution is t2 = 0.35 s

8 0
3 years ago
Anybody got an answer?
Anna11 [10]
For a regular n-agon with side length "s" and apothem "h", the area will be
.. A = n*(1/2)*s*h
Your area is
.. A = 10*(1/2)*(3.25 m)*(5 m) ≈ 81.3 m^2
8 0
3 years ago
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