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atroni [7]
3 years ago
13

Help Help Help Help Help Help HELPPPPP

Mathematics
2 answers:
tamaranim1 [39]3 years ago
8 0

Answer:

only D

Step-by-step explanation:

MA_775_DIABLO [31]3 years ago
4 0

Answer:

D

Step-by-step explanation:

You might be interested in
Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants. 37 x2 − 6x
Dvinal [7]

Answer:

\dfrac{A}{x}+\dfrac{B}{x-6}

Step-by-step explanation:

Given the function \dfrac{37}{x(x-6)}, to write the form of its partial fraction on decomposition, we will separate the two functions separated by an addition sign. The numerator of each function will be constants A and b and the denominator will be the individual factors of each function at the denominator. The partial fraction of the rational function is as shown below.

= \dfrac{37}{x(x-6)}\\\\= \dfrac{A}{x}+\dfrac{B}{x-6}

<em>Since we are not to solve for the constants, hence the partial fraction is </em>\dfrac{A}{x}+\dfrac{B}{x-6}

8 0
3 years ago
What’s the expression of this in simplest form
Ad libitum [116K]

Answer:

3/17

Step-by-step explanation:

3/4 divided by 4  1/4 is = to 3/4 divided by 17/4= 3/4 times 4/17 = 3/17

4 0
2 years ago
Read 2 more answers
How does 15.050 compare to 15.500
Varvara68 [4.7K]

15.050 < 15.500

notice that 500 is more than 50... so hope this helps a little bit !!!

:)

6 0
3 years ago
Read 2 more answers
A tank contains 180 gallons of water and 15 oz of salt. water containing a salt concentration of 17(1+15sint) oz/gal flows into
Stels [109]

Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}

and flows out at a rate of

\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))

Multiply both sides by the integrating factor, e^{t/180}, and rewrite the left side as the derivative of a product.

e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))

\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))

Integrate both sides with respect to t (integrate the right side by parts):

\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt

\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C

Solve for A(t) :

\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}

So,

\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}

Recall the angle-sum identity for cosine:

R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

R \cos(\theta) = -\dfrac{66,096,000}{32,401}

R \sin(\theta) = \dfrac{367,200}{32,401}

Recall the Pythagorean identity and definition of tangent,

\cos^2(x) + \sin^2(x) = 1

\tan(x) = \dfrac{\sin(x)}{\cos(x)}

Then

R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}

and

\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)

so we can rewrite A(t) as

\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}

and

24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}

which is to say, with amplitude

2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}

6 0
2 years ago
A town's recreation department is updating its parks. The parks must have slides and swings.
Fittoniya [83]
Number of swings = x
Number of slides = y

First condition:
number of slides must be more than 3 times number of swings
This means that: y > 3x

Second condition:
each swing cost 275$ and each slide costs 168$. They cannot spend more than 1250$. This means that the sum of costs of swings and slides should not exceed 1250$ (should be less than or equal to 1250$)
This means that: <span>275x+168y ≤ 1250

Based on this:
The correct choices are: b and d</span>
5 0
3 years ago
Read 2 more answers
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