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Ahat [919]
2 years ago
7

Ximena is typing a 2500 word essay. In 9 minutes she types 396 words, at this rate can Ximena type the essay in an hour?

Mathematics
1 answer:
sasho [114]2 years ago
8 0

Answer:

yes

Step-by-step explanation:

396÷ 9= 44

44×60= 2,640 > 2500

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How to slove this one?
disa [49]

Answer:

turning it I think.........

3 0
3 years ago
A store advertises 15% off an item that regularly sells for $300.
nataly862011 [7]

Answer:

The sale price of the item is  $ 255

Step-by-step explanation:

To determine the amount of discount ( d ) offered by the store, we calculate  

15 %  of the sale price. Thus:

d =  300  ×  15 ÷ 100

d  =  3 00  ×  15  ÷ 1 00  (take the 2 zeros out on both the hundred and 3 hundred)

d  =  3  ×  15

d  =  45

If the amount of discount is known, the sale price ( s ) of the item will be the regular price minus the discount. Hence:

s  =  300  −  45

s  =  255

Hope this helps you! If not sorry!

8 0
2 years ago
Solve this system by substitution (3 variables) brainliest answer goes to first answer and 8 points!
Natalka [10]

Answer:Solve this system by substitution (3 variables) brainliest answer goes to first answer and 8 points!

-x -y -z = -8

-4x +4y +5z = 7

2x+2z=4

Show steps please!

Step-by-step explanation: a-z=3-2=657u568959jb5j656856 thats the answer your welcome

4 0
3 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
3 years ago
A CHICKEN COOP HOLDS A TOTAL OF 10 HENS OR 20 CHICKS. IF 2/5THS OF THE HENS IN A FULL COOP ARE REMOVED, HOW MANY NEW CHICKS COUL
LiRa [457]

Answer:

8 new chicks can be fitted in the coop.

Step-by-step explanation:

A chicken coop holds 10 hens or 20 chicks.

That means space in the coop for 10 hens = space for 20 chicks

Or space for 1 hen = space for 2 chicks

Now \frac{2}{5}th of the hens were removed from the coop.

So, number of hens removed from the coop = \frac{2}{5}\times 10

= 4 hens

And space for 4 hens in the coop = space for the 8 chicks

Therefore, 8 new chicks can be fitted in the coop.

5 0
3 years ago
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