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Anestetic [448]
3 years ago
12

A circle in the xy-plane has a diameter with endpoints whose coordinates are negative 1 comma negative 3 and 7 comma 3. if the p

oint with coordinates 0 comma b lies on the circle and b is greater than 0, what is the value of b ?
Mathematics
2 answers:
DerKrebs [107]3 years ago
8 0

<span>First we look for the coordinates of the center of the circle. For this, we use the following formula:</span>

<span> ((x1 + x2) / 2, (y1 + y2) / 2)</span>

<span> Where,</span>

<span> (x1, y1) = (- 1, -3)</span>

<span> (x2, y2) = (7, 3)</span>

<span> Substituting values we have:</span>

<span> ((-1 + 7) / 2, (-3 + 3) / 2)</span>

<span> ((6) / 2, (0) / 2)</span>

<span> (3, 0)</span>

<span> We are now looking for the diameter of the circle. For this we use the formula of distance between points:</span>

<span> d = root ((x2-x1) ^ 2 + (y2-y1) ^ 2)</span>

<span> Substituting values:</span>

<span> d = root ((7 - (- 1)) ^ 2 + (3 - (- 3)) ^ 2)</span>

<span> d = 10</span>

<span> Then, the radius of the circle is:</span>

<span> r = d / 2 = 10/2</span>

<span> r = 5</span>

<span> The circle equation will be</span>

<span> (x-h) ^ 2 + (y-k) ^ 2 = r ^ 2</span>

<span> Where,</span>

<span> (h, k) = (3, 0)</span>

<span> r = 5</span>

<span> Substituting</span>

<span> (x-3) ^ 2 + (y-0) ^ 2 = 5 ^ 2</span>

<span><span> Substituting</span></span>

<span><span> (x, y) = (0, b)</span></span>

<span><span> (0-3) ^ 2 + (b-0) ^ 2 = 5 ^ 2</span></span>

<span><span> b = root (25-9)</span></span>

<span><span> b = root (16)</span></span>

<span><span> b = 4</span></span>

<span><span> Answer</span></span>

<span><span> the value of b is</span></span>

<span><span> <span>b = 4</span></span></span>


In-s [12.5K]3 years ago
5 0

Answer:

b = 4

Step-by-step explanation:

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Novosadov [1.4K]

Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math.  It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

y=Ce^{kt}

Filling in our formula with the 2 conditions we are given:

65=Ce^{10k}   and   85=Ce^{15k}

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k.  If

65=Ce^{10k} then

\frac{65}{e^{10k}}=C which, by exponential rules is the same as

C=65e^{-10k}

Since that value of C is the same as the value of C in the other equation, we sub it in:

85=65e^{-10k}(e^{15k})

Divide both sides by 65 and use the rules of exponents again to get

\frac{85}{65}=e^{-10k+15k} which simplifies down to

\frac{85}{65}=e^{5k}

Take the natural log of both sides to get

ln(\frac{85}{65})=5k

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

65=Ce^{10(.0536527973)}

which simplifies down to

65=Ce^{.536527973}

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

y=38.01038064e^{(.0536527973)(23)} which simplifies a bit to

y=38.01038064e^{1.234014338}

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

130.565

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3 years ago
A private airplane leaves an airport and flies due east at 192 km/hr. Two hours later, a jet leaves the same airport and flies d
elixir [45]
The answer is 0.5 hours.

A velocity (v) of an object is a distance (d) divided by time (t):
v = d ÷ t                         ⇒                  t = d ÷ v

It is given:
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v2 = 960 km/h
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Since they will travel the same distance before the jet overtakes the plane, we can say that d1 = d2 = d

Now, let's express t1 and t2:
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t2 = d/960

Therefore: d/192 - d/960 = 2
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3 years ago
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