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mojhsa [17]
3 years ago
12

Is the given point interior, exterior, or on the circle (x + 2)2 + (y - 3)2 = 81? P(8, 4)

Mathematics
1 answer:
Elodia [21]3 years ago
8 0
From the equation we see that the center of the circle is at (-2,3) and the radius is 9.

So using the distance formula we can see if the distance from the center to the point (8,4) is 9 units from the center of the circle...

d^2=(8--2)^2+(4-3)^2  and d^2=r^2=81 so

81=10^2+1^2

81=101  which is not true...

So the point (8,4) is √101≈10.05 units away from the center, which is greater than the radius of the circle.  

Thus the point lies outside or on the exterior of the circle...
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Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]
Zina [86]

Answer: Absolute minimum: f(-1) = -2\sqrt{6}

              Absolute maximum: f(\sqrt{12.5}) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = t\sqrt{25-t^{2}}

f'(t) = 1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)

f'(t) = \sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }

f'(t) = \frac{25-2t^{2}}{\sqrt{25-t^{2}} } = 0

For this function to be zero, only denominator must be zero:

25-2t^{2} = 0

t = ±\sqrt{2.5}

\sqrt{25-t^{2}} ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = -\sqrt{2.5} and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = t\sqrt{25-t^{2}}

f(-1) = -1\sqrt{25-(-1)^{2}}

f(-1) = -\sqrt{24}

f(-1) = -2\sqrt{6}

f(\sqrt{12.5}) = \sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}

f(\sqrt{12.5}) = 12.5

f(5) = 5\sqrt{25-5^{2}}

f(5) = 0

Therefore, absolute maximum is f(\sqrt{12.5}) = 12.5 and absolute minimum is

f(-1) = -2\sqrt{6}.

8 0
3 years ago
Find the measure of each unknown angle​
Aleks [24]
. The sum of angels in the triangle must be 180
Therefore
1) 180 - (40+85)= 55
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3 0
3 years ago
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What is the slope of a line that passes through the points (-3, 2) and (-6, 5)
oksano4ka [1.4K]

Answer:

3/2

Step-by-step explanation:

Y2-Y1

_____

X2-X1

= 5-2

____

-6-(-3)

=3/2

3 0
3 years ago
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Mason graphed the parent function of a quadratic equation. Then, he reflected the parabola over the x-axis and translated it sev
cluponka [151]

Answer:

  • y = - (x - h - 3)² - k + 7

Step-by-step explanation:

<u>Let the patent function be:</u>

  • y = (x - h)² + k in the vertex form

<u>Then transformation steps are:</u>

1. Reflection over x- axis, y' = -y

  • y' = - (x - h)² - k

2. Translation 7 units up and 3 units to the right, y'' = y' + 7, x' = x - 3

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7 0
3 years ago
Find the distance between each pair of points.
BigorU [14]

Answer:

Distance between the points (-3,1) and (1, -2),

\sqrt{ {(1 + 3)}^{2} +  {( - 2 - 1)}^{2}  }

=  \sqrt{ {(4)}^{2} +  {( - 3)}^{2}  }

=  \sqrt{16 + 9}  =  \sqrt{25}  = 5

Option C

8 0
4 years ago
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