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Readme [11.4K]
3 years ago
7

Siny dy/dx= cosy(1-xcosy)pls i need it fast​

Mathematics
1 answer:
crimeas [40]3 years ago
5 0

Answer:

sec y = x+1 + ce x ; c is constant

Hope it helps....

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Ksenya-84 [330]
3rd one trust me bro
8 0
3 years ago
Select all of the following expressions that are equivalent to 0.25. (0.5)2 (-0.2) + (-0.05) |-3 + 2.75| -(-0.25) 2x - 1.75x
stiv31 [10]
The solutions for the following expressions are:
1. 0.5*2 = 1
2. -0.2+-0.05= -0.25
3. absolute value of (-3+2.75) = 0.25
4. -(-0.25) = 0.25
<span>5. 2x-1.75x = 0.25x</span><span>


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3 0
4 years ago
During the 2012-2013 NBA season, Jamal Crawford playing for the Los Angeles Clippers, had a free throw percentage of 0.871. Usin
yanalaym [24]

Answer:

When Jamal Crawford comes to the free throw line, in the 2012-2013 NBA season, he would make 871 free throws out of 1000.  Therefore, his percentage of making a free throw would by %87.1

8 0
4 years ago
George drove 168 miles in 4 hours. If this rate continues, how many miles can he drive in 7 hours?
Nataly [62]
Find the unit rate. He drove 168 miles in 4 hours.

168/4= 42

He drives 42 Mph. 

Multiply that by 7.

42*7= 294

He drives 294 in 7 hours.

I hope this helps!
~kaikers
6 0
4 years ago
Read 2 more answers
Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of
balu736 [363]

Denote the circle of radius a by C. C is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of C (call it D):

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy

=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy

=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy

D is a circle of radius a, so we can write the double integral in polar coordinates as

\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta

a. For a=1, we have

\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}

b. Let I(a) denote the integral with unknown parameter a,

I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta

By the fundamental theorem of calculus,

I'(a)=12\pi(9a-a^3)

I(a) has critical points when

12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3

If a=0, then line integral is 0, so we ignore that critical point. For the other two, we would find I(\pm3)=243\pi.

8 0
3 years ago
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