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3 years ago

Please help NOWW! :D

Mathematics

2 answers:

cupoosta [38]3 years ago

8 0

I really need a brainlest can you help me out and if you tell me where the rectangle is I will edit my answer

Delvig [45]3 years ago

6 0

Answer: what rectangle???

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Which of the following are reasons why a bank may establish a multinational operation?

Rashid [163]

Answer:

multinational banking operations help a bank prevent the erosion of its traveler's check, tourist, and foreign business markets from foreign bank competition.

mark me brainliest.

3 0

2 years ago

Number of Certified Organic Farms in the United States, 2001–2008

Vinil7 [7]

Answer:

ŷ = 739.49X + 4876.43

y = 6755.98 - 388.24x + 125.30x²

y = 5428.98(1.09)^x

B.)

Linear:

ŷ = 739.49(9) + 4876.43

y = 11531.8

Year 2010 ; x = 10

y = 739.49(10) + 4876.43

y = 12271.3

Year 2011 ; x = 11

y = 739.49(11) + 4876.43

y = 13010.8

Quadratic :

Year 2009 ; x = 9

y = 6755.98 - 388.24(9) + 125.30(9^2)

y = 13411.1

Year 2010 ; x = 10

y = 6755.98 - 388.24(10) + 125.30(10^2)

y = 15403.6

Year 2011 ; x = 11

y = 6755.98 - 388.24(11) + 125.30(11^2)

y = 17646.6

Exponential:

Year 2009 ; x = 9

y = 5428.98(1.09)^9

y = 11791.2

Year 2010 ; x = 10

y = 5428.98(1.09)^10

y = 12852.4

Year 2011 ; x = 11

y = 5428.98(1.09)^11

y = 14009.1

Step-by-step explanation:

X :

6231

6574

7237

7211

7701

8581

10302

11796

Using the online linear regression calculator :

The linear trend :

ŷ = 739.49X + 4876.43

Where x = year

With 2006 representing 1 ; and so on

Slope = m = 739.49

Intercept (c) = 4876.43

y = predicted variable

The quadratic model:

General form:

y = A + Bx + Cx²

y = 6755.98 - 388.24x + 125.30x²

The exponential model:

y = AB^x

y = 5428.98(1.09)^x

B.) Next three years :

Year 2009 ; x = 9

Year 2010 ; x = 10

Year 2011 ; x = 11

Linear:

ŷ = 739.49(9) + 4876.43

y = 11531.8

Year 2010 ; x = 10

y = 739.49(10) + 4876.43

y = 12271.3

Year 2011 ; x = 11

y = 739.49(11) + 4876.43

y = 13010.8

Quadratic :

Year 2009 ; x = 9

y = 6755.98 - 388.24(9) + 125.30(9^2)

y = 13411.1

Year 2010 ; x = 10

y = 6755.98 - 388.24(10) + 125.30(10^2)

y = 15403.6

Year 2011 ; x = 11

y = 6755.98 - 388.24(11) + 125.30(11^2)

y = 17646.6

Exponential:

Year 2009 ; x = 9

y = 5428.98(1.09)^9

y = 11791.2

Year 2010 ; x = 10

y = 5428.98(1.09)^10

y = 12852.4

Year 2011 ; x = 11

y = 5428.98(1.09)^11

y = 14009.1

6 0

2 years ago

What is the solution to the system of equations below?<br> y = 3x + 9<br> y = -3x + 12

Elena L [17]

Answer:

Step-by-step explanation:

y = 3x + 9

y = -3x + 12

y=21

7 0

2 years ago

Read 2 more answers

5 -2 (4a + 1) + 3a = 13 what is (a) equal to a) - 6/5 b) - 2 c) 2 d) 6/5

Sidana [21]

The correct answer is a= -2

4 0

2 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