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xxMikexx [17]
3 years ago
11

Match the following exponential functions with their "b" value.

Mathematics
1 answer:
marishachu [46]3 years ago
4 0

9514 1404 393

Answer:

  in order by function: 0.6, 8, 7/3, 2, 1/2

Step-by-step explanation:

The functions are given in the form ...

  f(x) = a·b^x

The "b" value is the value immediately to the left of the exponent. (It's not rocket science; it's pattern matching.) Note that the minus sign in g(x) is part of 'a', not part of 'b'.

From the top-down, the functions listed on the left have the b-values shown above.

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Sanjay sold 8kg of sugar at Rs 28 per kg and earned a profit of Rs 40 What was the cost price per kg
JulijaS [17]

Answer:

23

Step-by-step explanation:

8 kg * 28 per kg = Rs. 224

Profit = Rs. 40

CP = SP - Profit

224 - 40

184

184 / 8

23

4 0
2 years ago
Given the diagram below, what is cos 45
Mamont248 [21]
Hey

Your answer is pretty simple

1/√2
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Which of the following algebraic expressions correctly represent the phrase fifteen less than twice a number
Vlad1618 [11]

Answer:

15-(2×X)

Step-by-step explanation:

hope this is what you need

8 0
3 years ago
Read 2 more answers
The square root of 7 lies between what two numbers
ExtremeBDS [4]

The square root of 7 would like between 2 and 3. This is because 2 x 2 is 4's square root and 3 x 3 is nine's square root. Seven is between four and nine.

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3 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
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