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3 years ago

Solve the solution. How many solutions does it have? 4(2x-3)=2(4x-6)

Mathematics

1 answer:

sdas [7]3 years ago

7 0

Answer:

No solutions

Step-by-step explanation:

4(2x-3)=2(4x-6)

$4(2x-3)=2 \times 2(2x-3) \\ \\ 4(2x-3)=4(2x-3)$

Same components are present on both sides of equal sign, therefore NO SOLUTIONS are possible.

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Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

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3 years ago

when you reverse the digits in a certain two-diget number you increase it value by 27. What is the number if the sum of its digi

Fynjy0 [20]

The number is = 10x+y
if reverse the digits, the number is: 10y+x
The sum fo its digits is = x+y=5
I suggest this system of equations:
(10y+x)-(10x+y)=27 ⇒9y-9x=27
x+y=5
We can solve this system of equations by equaliztion method
x=5-y

9y-9(5-y)=27
9y-45+9y=27
18y=27+45
18y=72
y=72/18
y=4

x=5-y
x=5-4=1

The number is: 10x+y=10(1)+4=14

Answer: the number is 14

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2 years ago

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Solve the system of linear equations below. Enter your solution as an ordered pair (x,y). Do not include spaces in your answer..

kotykmax [81]

Y=3
4x+y=19
4x+3=19
4x=19-3
4x=16
x=4

pair is 4,3

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2 years ago

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Which of the following is positive? -(-31), -(31), -31, 0-31

BartSMP [9]

It would be -(-31) because a negative and negative is a positive

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3 years ago

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How many Solutions does this system have? (1 point)

mixas84 [53]

The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$

As $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ , so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

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2 years ago