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MArishka [77]
2 years ago
10

One summer, an ice cream truck driver sold 381 Popsicles and 76 ice creams. How many treats did the driver sell in all?

Mathematics
1 answer:
vazorg [7]2 years ago
8 0
The answer is 457 because when you add 381 and 76 together you get 457
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Find the value of given expression<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B1221%20%5Ctimes%201221%7D%20" id="TexFo
densk [106]

Answer:

1221\sqrt{x}

Step-by-step explanation:

1221 x 1221 = 1490841

square root of 1490841 is 1221

8 0
2 years ago
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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
tresset_1 [31]

Because I've gone ahead with trying to parameterize S directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over S straight away, let's close off the hemisphere with the disk D of radius 9 centered at the origin and coincident with the plane y=0. Then by the divergence theorem, since the region S\cup D is closed, we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV

where R is the interior of S\cup D. \vec F has divergence

\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z

so the flux over the closed region is

\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0

\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}

Parameterize D by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k

with 0\le u\le9 and 0\le v\le2\pi. Take the normal vector to D to be

\vec s_u\times\vec s_v=-u\,\vec\jmath

Then the flux of \vec F across S is

\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0

\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}

8 0
3 years ago
Please help i beg i never get these right
Anuta_ua [19.1K]

Answer:

\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}

Step-by-step explanation:

Given

\sin(105^o)

Required

Solve

Using sine rule, we have:

\sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A)

This gives:

\sin(105^o) = \sin(60 + 45)

So, we have:

\sin(60 + 45) = \sin(60)\cos(45) + \sin(45)\cos(60)

In radical forms, we have:

\sin(60 + 45) = \frac{\sqrt 3}{2} * \frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} * \frac{1}{2}

\sin(60 + 45) = \frac{\sqrt 6}{4} + \frac{\sqrt 2}{4}

Take LCM

\sin(60 + 45) = \frac{\sqrt 6 + \sqrt 2}{4}

Rewrite as:

\sin(60 + 45) = \frac{\sqrt 2 + \sqrt 6}{4}

Hence:

\sin(105) = \frac{\sqrt 2 + \sqrt 6}{4}

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2 years ago
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morpeh [17]

Answer:

C.

Step-by-step explanation:

y = 6x

if x = 1 , y = 6

if x = 2 , y = 12

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2 years ago
Show all work to identify the asymptotes and zero of the function f of x equals 5 x over quantity x squared minus 25.
Lorico [155]

Answer:

  • asymptotes: x = -5, x = 5
  • zero: x = 0

Step-by-step explanation:

The function of interest is ...

  f(x)=\dfrac{5x}{x^2-25}=\dfrac{5x}{(x-5)(x+5)}

The asymptotes are found where the denominator is zero. It will be zero when either factor is zero, so at x = 5 and x = -5

__

The zeros are found where the numerator is zero. It will be zero for x = 0.

The asymptotes are x=-5, x=5; the zero is x=0.

4 0
2 years ago
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