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3 years ago

I need help with this I need love

Mathematics

2 answers:

Greeley [361]3 years ago

7 0

Answer:

A tourism/tourist product can be defined as the sum of the physical and psychological satisfaction it provides to tourists, during their 'traveling and sojourn' en route at the destinations. ... Thus, whatever the natural and man-made resources, services brought about the consumption of tourists is called tourism products.

solniwko [45]3 years ago

3 0

8:00? it means 20:00

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What is 2.05 +4.77 ?

Effectus [21]

The answer is 6.82

2 + 4 = 6

.05 + .77 = .82

6 + .82 = 6.82

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2 years ago

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Which of the following could NOT be the lengths of the sides of a triangle?

Furkat [3]

Given:

Sides of triangles in the options.

To find:

Which could NOT be the lengths of the sides of a triangle.

Solution:

Condition for triangle:

Sum of two smaller sides of a triangle must be greater than the longest side.

In option A,

$5+5=10>5$

Sides 5 in, 5 in, 5 in are the lengths of the sides of a triangle.

In option B,

$10+15=25>20$

Sides 10 cm, 15 cm, 20 cm are the lengths of the sides of a triangle.

In option C,

$3+4=7>5$

Sides 3 in, 4 in, 5 in are the lengths of the sides of a triangle.

In option D,

$8+5=13$

Since, the sum of two smaller sides is less than the longest side, therefore the sides 8 ft, 15 ft, 5 ft are not the lengths of the sides of a triangle.

Therefore, the correct option is D.

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3 years ago

Jean Claude took out a $3,500 installment loan at 11% for 12 months. His monthly payments were $309.34. After seven payments, th

bezimeni [28]

$27.86 is the answer

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3 years ago

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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

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

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2 years ago

Find the coordinates of the circle and the radius y²+2x+x²-24y= -120

riadik2000 [5.3K]

We know that
the equation of a circle is
(x-h+(y-k)²=r²
where (k,k)----> is the center
and
r----> is the radius
y²+2x+x²-24y= -120

Group terms that contain the same variable
(y²-24y)+(x²+2x)=-120

Complete the square twice. Remember to balance the equation by adding the same constants to each side
(y²-24y+144)+(x²+2x+1)=-120+144+1

<span>Rewrite as perfect squares</span>

(y-12)²+(x+1)²=25

(x+1)²+(y-12)²=5²

the answer is
the coordinates of the center are (-1,12)
the radius is 5 units

3 0

3 years ago