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MakcuM [25]
3 years ago
8

100 points! :) How do you find the equation of a line given two points? Please explain step-by-step!

Mathematics
2 answers:
monitta3 years ago
6 0

Answer:

-2+4=2, 1+8=9

Hope I helped

geniusboy [140]3 years ago
5 0

Answer:

1,9

Step-by-step explanation:

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For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec
telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

4sin(x^2) = 0

sin(x^2) = 0

x^2 = n\pi where n = 0, 1, 2,3, 4, 5,...

x = \sqrt(n\pi)

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0

xcos(x^2) = 0

x = 0 or

cos(x^2) = 0

x^2 = \frac{\pi}{2} + n\pi where n = 0, 1, 2, 3

x = \sqrt{\pi(n+1/2)}

Since we are restricting x between 0 and 3 we can stop at n =  2

So our critical points are at

x = 1.25, y = f(1.25) = 4sin(1.25^2) = 4

x = 2.17 , y = f(2.17) = 4sin(2.17^2) = -4

x = 2.8, y = f(2.8) = 4sin(2.8^2) = 4

For the inflection point, we can take the 2nd derivative and set it to 0

f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0

cos(x^2) = 2x^2sin(x^2)

tan(x^2) = \frac{1}{2x^2}

We can solve this numerically to get the inflection points are at

x = 0.81, y = f(0.81) = 4sin(0.81^2) = 2.44

x = 1.81, y = f(1.81) = 4sin(1.81^2) = -0.54

x = 2.52, y = f(2.52) = 4sin(2.52^2) = 0.27

3 0
3 years ago
The word that is associated with doing something to help protect something is —
tamaranim1 [39]
The answer is c i’m pretty sure
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Answer:

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Step-by-step explanation:

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7 0
3 years ago
Racionalize o denominador da equação abaixo:<br> 5 + 3 ²√5 / ²√5
GrogVix [38]
I hope this helps

\cfrac{5+3 \sqrt{5} }{ \sqrt{5} } =\cfrac{(5+3 \sqrt{5}) \sqrt{5}  }{ \sqrt{5}* \sqrt{5}  }= \cfrac{5 \sqrt{5}+15 }{5}  = \cfrac{5 (\sqrt{5}+3) }{5}  = \sqrt{5}+3
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n200080 [17]

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6 0
2 years ago
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