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mylen [45]
3 years ago
12

Which of the following equations is paralllel to the line y+3x+1 and passes through the point (6,-3)

Mathematics
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

y=3x-21

Step-by-step explanation:

a line is parallel if the slopes of the lines are the same

so that means that out line should look something like this

y=3x+b

then, we use our point given to find b

-3=18+b

b=-21

the equation of out line is y=3x-21

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You buy a watch for $60. There is a 6% sales tax. What is your total cost for the<br> watch?
lara [203]

Answer:

$63.3

Step-by-step explanation:

You buy a 60 dollar watch.

60

There is a 6 percent tax. When adding tax, multiply the 6 percent by the original amount.

60 x 0.06

3.6

Add the tax to the original amount.

60 + 3.6 = 63.3

So, your total cost is 63.3 dollars.

4 0
2 years ago
For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f
butalik [34]

\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k

Let

\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k

The curl is

\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)

where \partial_\xi denotes the partial derivative operator with respect to \xi. Recall that

\vec\imath\times\vec\jmath=\vec k

\vec\jmath\times\vec k=\vec i

\vec k\times\vec\imath=\vec\jmath

and that for any two vectors \vec a and \vec b, \vec a\times\vec b=-\vec b\times\vec a, and \vec a\times\vec a=\vec0.

The cross product reduces to

\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

\nabla\times\vec f=\vec0

which means \vec f is indeed conservative and we can find f.

Integrate both sides of

\dfrac{\partial f}{\partial y}=2xze^{2xyz}

with respect to y and

\implies f(x,y,z)=e^{2xyz}+g(x,z)

Differentiate both sides with respect to x and

\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}

2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}

4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}

\implies g(x,z)=4\sin(xz^2)+h(z)

Now

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)

and differentiating with respect to z gives

\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}

2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0

\implies h(z)=C

for some constant C. So

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C

3 0
3 years ago
ALG TWO HELP ASAP?
Alexxx [7]

Answer:

Therefore the period of the sinusoidal wave = 3.14.

The amplitude of the function = 2.

Step-by-step explanation:

The period of a sinusiodal wave is given by the length of x axis which covers one full cycle of the wave, which one positive half-cycle and one negative half cycle.

From the graph we can see  that one of the cycles starts at -3.14 / 4 and ends at 3.14 \times 3/4.

Therefore we can sutract the start point value from the end point value to get the period

Therefore period  = (3.14 \times 3/4) - (-3.14 / 4) = (3.14 \times 3/4) + (3.14 / 4) = 3.14

Therefore the period of the sinusoidal wave = 3.14.

The amplitude of the function = 2.

6 0
3 years ago
Write the expression for 5 minus q
levacccp [35]

Answer:

5 - q

Step-by-step explanation:

the term minus indicates a sign of subtraction, which looks like this

-

Therefore, the expression is 5 - q

8 0
2 years ago
What is the standard form of five and three hundred fifty two thousands
suter [353]
The number "seven hundred and four thousand, six hundred and twenty" is written as 704,620 in numeric or standard form.<span> In converting a number expressed in words to numerals, it is important to write each digit in its proper place value.</span>
7 0
3 years ago
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