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3 years ago

What is the radius of a circle whose equation is x2 + (y – 8)2 = 25?

1 answer:

6 0

The formula for a circle is (x - h)^2 + (y - k)^2 = r^2
Where (h, k) is the center and r is the radius.
x^2 + (y - 8)^2 = 25
25 = r^2
So the radius is sqrt 25 or 5.
Radius = 5

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I am completely stumped...

Nana76 [90]

Answer:

x=25 $\sqrt{2}$

Step-by-step explanation:

the hypotenuse is equal to a leg times the square root of 2. which means to find the leg with the hyp given. you take $\frac{50}{\sqrt{2} }$ which then translates to 50 root 2/2=x therefore 25 $\sqrt{2}$

7 0

3 years ago

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

On a map, 3 inches represents 50 miles. Which proportion can be used to find the actual distance represented by 5 inches on the

asambeis [7]

The correct proportion that can be used to find the actual distance represented by 5 inches on the map is start fraction over 50 end fraction=start fraction 50 over 5 end fraction which is 3/50=5/x.

Given that 3 inches on a map represents 50 miles.

We are told to find the proportion which can be used to represent 5 inches.

Multiplication is basically finding the product of two numbers but it can be used to convert the units also.

If 3 inches represents 50 miles then,

1 mile=50/3

Multiply both sides by 5.

5 miles=5*50/3

If we see the options then we will get that the most appropriate and correct option which represents 5 inches is that start fraction over 50 end fraction=start fraction 50 over 5 end fraction which is 3/50=5/x.

Hence the correct proportion that can be used to find the actual distance represented by 5 inches on the map is start fraction over 50 end fraction=start fraction 50 over 5 end fraction which is 3/50=5/x.

Learn more about fraction at brainly.com/question/78672

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5 0

2 years ago

Let f(p) be the average number of days a house stays on the market before being sold for price p in $1,000s.

vampirchik [111]

The correct answer is:

This is the average number of days the house stayed on the market before being sold for $150,000.

Explanation:

f(p) is defined as the average number of days a house stays on the market before being sold for price p (given in $1000).

We want f(150); this means p=150. Since p is in thousands of dollars, this means the price of the house was $150,000.

This means f(150) is the average number of days the house stayed on the market before being sold for $150,000.

4 0

4 years ago

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kenny6666 [7]

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8 0

2 years ago

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