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Tpy6a [65]
3 years ago
6

Find the equation of the parabola with vertex (1,3) and focus (1/2,3).

Mathematics
1 answer:
daser333 [38]3 years ago
4 0

Answer:

A (y-3)^2 = -2(x-1) is the correct choice :)


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Question 1.:
klasskru [66]

Answer:

-7 is bigger

-9 is bigger

-6 is bigger

7 is bigger

I think.....

-9,-8,-7,-6,-4

-9,-7,-6,4,7

-2-1,0,4,9

-8,-2,3,5,8 ?

I guess I didn't understand the question

8 0
3 years ago
Lilli jogs 12 mile in 110 hour What is Lilli's rate in miles per hour?
Verizon [17]
The problem requires us to find the rate or speed of Lilli in miles per hour. Therefore, we just have to divide the given data.

12/110 = 0.10909.... Therefore, Lilli's speed or rate is 0.11 mi/hr.
5 0
4 years ago
3(c+5)=12. Distributive property
avanturin [10]

Answer:

c = -1

Step-by-step explanation:

3(c+5)=12

3 × c = 3c

3 × 5 = 15

3c + 15 = 12

<u>      - 15   -15</u>

   3c = -3

c = -3 ÷ 3 = -1

Check:

3(-1+5)=12

-3 + 15 = 12

6 0
3 years ago
Read 2 more answers
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
What is the percent of decrease from 100 to 49?<br><br> Write your answer using a percent sign (%).
slavikrds [6]

Answer:

Step-by-step explanation:

Decrease = 100 - 49 = 51

Percentage of decrease =

=\frac{51}{100}*100

= 51%

6 0
3 years ago
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