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Ray Of Light [21]
2 years ago
14

The vertex if the parabola below is at the point (2,4) and the point (3,6) is on the parabola. What is the equations the parabol

a
Mathematics
2 answers:
Sloan [31]2 years ago
4 0
I think the equation is 2(x-2)^2+4
3241004551 [841]2 years ago
4 0

Answer:

y = 2(x-2)^2+4  is the equations the parabola

Step-by-step explanation:

The standard equation of parabola is given by:

y = a(x-h)^2+k           ....[1]

where,

a is the non zero real number and

(h, k) is the vertex of the parabola.

As per the statement:

The vertex if the parabola below is at the point (2,4)

⇒(h, k) = (2, 4)

⇒h =2 and k =4

Substitute these in [1] we have;

y = a(x-2)^2+4

Since, the point (3,6) is on the parabola

⇒x = 3 and y = 6

then;

6 = a(3-2)^2+4

6 = a+4

Subtract 4 from both sides we have:

2= a

or

a = 2

⇒y = 2(x-2)^2+4

Therefore, the equations the parabola is, y = 2(x-2)^2+4

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Evaluate and match each expression on the left to its value on the right, when x = 9 and y = 6.
kolezko [41]
Answers: 12-x=3 | x+3y=27 | 4y-10=14 | 1/3xy=18

Explanation: x=9 and y=6
12-(9)=3 | (9)+3(6)=27 | 4(6)-10=14 | 1/3(9)(6)=18

5 0
3 years ago
What is 230% of 90?<br> A 207<br> B 237<br> C 320<br> D 227
lilavasa [31]

Answer:

207 or option a is the answer

7 0
2 years ago
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Emergency! I don't get this really well. Please help!<br><br>-(8-2n)=8-2(12-3n)
Naddik [55]

Answer:

n = 2

Step-by-step explanation:

2n - 8 = -2 (-3n + 12) + 8

2n - 8 = -(-6n + 24) + 8

2n - 8 = 6n - 24 + 8

2n - 8 = 6n - 16

-4n = -8

Divide each side by 4

n = 2

3 0
3 years ago
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PLS HELP
mamaluj [8]
6... I think.
I hope that this is right
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3 years ago
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Your swimming pool containing 60,000 gal of water has been contaminated by 6 kg of a nontoxic dye that leaves a swimmer's skin a
Paul [167]

[a] Dye is removed from the pool at a rate of

(250 gal/min) * (<em>q</em>/60,000 g/gal) = -<em>q</em>/240 g/gal

where <em>q</em> denotes the amount of dye in the pool at time <em>t</em>. Clean water is pumped back into the pool, so no dye is being re-added.

So the net rate of change of the amount of dye in the pool is given by the differential equation,

\dfrac{\mathrm dq}{\mathrm dt}=-\dfrac{q(t)}{240}

with the intial value, <em>q</em>(0) = 6000 g (or 6 kg).

[b] The ODE above is separable as

\dfrac{\mathrm dq}q=-\dfrac{\mathrm dt}{240}

Integrate both sides to get

\ln|q|=-\dfrac t{240}+C

e^{\ln|q|}=e^{-t/240+C}

\implies q(t)=e^{-t/240+C}=e^{-t/240}e^C=Ce^{-t/240}

Now plug in the initial condition:

6000=Ce^0\implies C=6000

so the particular solution to the IVP is

q(t)=6000e^{-t/240}

[c] The acceptable concentration of the dy is 0.03 g/gal, which in a pool containing 60,000 gal of water corresponds to

(0.03 g/gal) * (60,000 gal) = 1800 g = 1.8 kg

of dye. Find the time <em>t</em> when this occurs:

1800=6000e^{-t/240}\implies0.3=e^{-t/240}

\implies\ln0.3=-\dfrac t{240}

\implies t=-240\ln0.3\approx288.953

so the amount of dye in the pool is within the acceptable tolerance after about 289 min have passes, or about 4.82 hrs. So no, the filtration system is not up to the task.

8 0
3 years ago
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