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2 years ago

10

Give the center and radius of the circle described by the equation and graph the

2 answers:

MrRissso [65]2 years ago

5 0

Answer:

1234567u8i9o0

Paraphin [41]2 years ago

3 0

Answer:

I don't know the exact answer but, I had a problem like this the other day. This is how to find it out. I am just in a rush so I have to go if not I would help you figure it out.

Step-by-step explanation:

Write the equation of each circle.

1. center at (9, 0), radius 5

SOLUTION:

ANSWER:

2. center at (3, 1), diameter 14

SOLUTION:

Since the radius is half the diameter, r = (14) or 7.

ANSWER:

3. center at origin, passes through (2,2)

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I need help with my math!

pantera1 [17]

Answer:

there are a total of 20 boys and 30 girls in the choir of 50 students

Step-by-step explanation:

Scale the ratio 2:3. so that the numbers on either side sum to 50 . 2+3=5 , and 50÷5=10 , so we scale our ratio by a factor of 10 . So there are a total of 20 boys and 30 girls in the choir of 50 students

3 0

3 years ago

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh

fredd [130]

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

6 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F. Af

fgiga [73]

Answer:

53.3324

Step-by-step explanation:

given that a thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F.

By Newton law of cooling we have

T(t) = $T+(T_0-T)e^{-kt}$

where T (t) is temperature at time t,T =surrounding temperature = 40, T0 =70 = initial temperature

After half minute thermometer reads 60° F. Using this we can find k

$T(0,5) = 40+(70-40)e^{-k/2} = 60\\e^{-k/2} =2/3\\-k/2 = -0.4055\\k = 0.8110$

So equation is

$T(t) = 40+(30)e^{-0.8110t}\\$

When t=1,

we get

$T(1) = 40+(30)e^{-0.8110}\\\\=53.3324$

5 0

3 years ago

Find the focus of the parabola whose vertex is at<br> the origin and whose directrix is x = 3.

defon

Answer:

Step-by-step explanation:

Vertex is halfway between focus and directrix, so focus is (-3,0).

6 0

2 years ago

What is the rate of change of the linear function that passes through the points (1, 0) and (0, -3)

melisa1 [442]

Answer:

3

Step-by-step explanation:

To find the rate of change we: divide the change in y (output) by the change in x (input).

A = (-3 - 0) / (0 - 1) = -3 / -1 = 3

:)

6 0

2 years ago