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2 years ago

Helllo fellow class maters

Mathematics

2 answers:

mr_godi [17]2 years ago

5 0

Answer:

Step-by-step explanation:

bixtya [17]2 years ago

4 0

Answer:

hello

Step-by-step explanation:

how are u :)

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Help with #9 please?

nekit [7.7K]

We're given a parabola

$y=x^2-4$

and a line

$y = 2x+b$

The zeros of the parabola are

$0=x^2-4$

$4=x^2$

$x=\pm 2$

We see that (-2,0) is on the line as well:

$y = 2x+b$

$0 = 2(-2)+b$

$b=4$

Choice D

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2 years ago

PQ is tangent to the circle at C. In the circle the measure of arc BC=85. Find the measure of angle BCP

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kap26 [50]

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Klio2033 [76]

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2 years ago

Read 2 more answers

evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

meriva

We have

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt$

and

$\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k$

$\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k$

so the line integral is equilvalent to

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}$

7 0

2 years ago

Helllo fellow class maters​

Helllo fellow class maters