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Alexxx [7]
3 years ago
6

Amy, Dan, and mike are competing in a cooking competition. They all used different amounts of milk from a container holding 6 ga

llons of milk. Amy used 13
Mathematics
1 answer:
Flauer [41]3 years ago
4 0

Answer:

2 1/6 containers of milk, if that's what you're asking.

Step-by-step explanation:

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Aleksandr-060686 [28]
For one of the triangular sides, 2.6 x 4 = 10.4 divided by 2 = 5.2
Other triangle; 2.6 x 3 = 7.8 /2 = 3.9
Rectangular side in front; 6.5 x 3 = 19.5
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Base; 6.5 x 2.6 = 16.9
Answer: For the surface area it is 74.6 (SUPER SORRY IF THIS ISNT RIGHT OMG)
5 0
4 years ago
What is the description of the pattern in the sequence?
Ronch [10]
Multiplying by 2, the next one would be 160
8 0
3 years ago
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How are the solutions to the inequality -2x ≥ 10 different from the solutions to -2x > 10?
CaHeK987 [17]

Answer:

First lets find the solutions to each inequality.

-2x\geq \\10 and -2x>10 (divide both sides by -2 to solve)

x\leq-5 and x<-5

x\leq-5 tell us that x could be -5 or less.

x<-5 tells us that x could be -6 or less.

The first one is less than or equal to which tells you that there is a possibility that the number is shows is could be an answer

Hope this helps ;)

4 0
4 years ago
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
4 years ago
Can y’all help with these to answer?
Soloha48 [4]

Answer:

1) 9:14

2)4:1

Step-by-step explanation:

You have to have all of the teachers compared to all of the adults (teachers included)

You have to simplify 12:3 and you get 4:1.

3 0
3 years ago
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