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3 years ago

Help pls its iready ill give brainiest

Mathematics

2 answers:

Nady [450]3 years ago

8 0

C=2pi2
c=12.56

so the last one

ladessa [460]3 years ago

5 0

Answer:

I think that it is the second one

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I need help with this problem right away please.

irga5000 [103]

Answer:

y= -1/9(x-1)^2 +2

Step-by-step explanation:

The vertex is at (1,2) and another point is at (-2,1).

We know the vertex form of a parabola is

y= a(x-h)^2 +k where (h,k) is the vertex

Substituting the vertex in

y= a(x-1)^2 +2

We have another point, (-2,1)

Substitue this in with x=-2 and y =1. This will let us find a

1 = a(-2-1)^2 +2

1 = a (-3)^2 +2

1 = a*9 +2

Subtract 2 from each side

1-2 = 9a +2-2

-1 = 9a

Divide by 9

-1/9 = 9a/9

-1/9 =a

Putting this back into the equation

y= -1/9(x-1)^2 +2

7 0

3 years ago

How to solve the equation (x+k) = (x+k-1)(x+k)?

Brilliant_brown [7]

This solution might become quite complex.
Please try to keep up with me, and I'll go slowly:

You said       (x + k)   = (x + k - 1) (x + k)

Divide each side by (x+k):    1     = (x + k - 1)

Add 1 to each side: 2       = (x + k )

Subtract 'k' from each side:    2 - k = x

3 0

3 years ago

Read 2 more answers

Which of the following relations represent a function? A. {(-1,-1),(0,0),(2,2),(5,5)}. B. None of these. C. {(0,3),(0,-3),(-3,0)

Cerrena [4.2K]

Answer:

Step-by-step explanation:

In a list of ordered pairs, a relation that is a function will not have any repeated x-values.

A -- a function (x-values are {-1, 0, 2, 5} with no repeats)

B -- irrelevant

C -- not a function (x-values are {0, 0, -3, 3} with 0 being repeated)

D -- not a function (x-values are {-2, -1, -2, 2} with -2 being repeated)

3 0

3 years ago

Least common multiple of 8, 12 and 30

Fittoniya [83]

Least common multiple of 8, 12, and 30 is 2

6 0

3 years ago

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) * (q/60,000 g/gal) = -q/240 g/gal

where q denotes the amount of dye in the pool at time t. 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, q(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 t 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

Read 2 more answers