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3 years ago

Which of the following is an extraneous solution of -3 x-2= x+2?

Mathematics

2 answers:

Elina [12.6K]3 years ago

6 0

X = (-1) because you add two to both sides the subtract the x and you get -4x=4 they you divid by -4 and get -1

Svetlanka [38]3 years ago

5 0

Answer:

x =(-1)

Step-by-step explanation:

There you go, buddy!

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Expand the following expression: 4 (2k + j)

Nadya [2.5K]

Answer:

4(2k + j) = 8k + 4j

Step-by-step explanation:

4(2k + j)

Distribute in the 4.

4 * 2k + 4 * j

Multiply 4 and 2k.

8k + 4 * j

Multiply 4 and j.

8k + 4j

3 0

2 years ago

Read 2 more answers

Simplity

Vikentia [17]

Answer:

5 ^ (5/6)

Step-by-step explanation:

5 ^ (1/2) * 5 ^ (1/3)

We know that a^b * a^ c = a ^ (b+c)

5 ^ (1/2 + 1/3)

5^ (3/6+2/6)

5^ (5/6)

8 0

3 years ago

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While vacationing, Rosalie sees a mountain and its reflection on the still water of a lake. She draws the mountain and its refle

Greeley [361]

Answer:

I would say (2,-9) is her projection

6 0

3 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago

A 64 ounce bottle of organic orange juice costs $14.72. What is the price

olasank [31]

A 64 ounce bottle of organic orange juice costs $14.72. What is the price

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8 0

2 years ago