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3 years ago

Given the equation 2 Square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

2 answers:

8 0

2 square root (x - 5) = 2. 2^2 (x -5) = 2^2. 4(x - 5) = 4. 4x - 20 = 4. 4x = 24. x = 6. This solution is not extraneous, because extraneous solutions emerge from solving the problem but are not actually valid solutions for the initial problem. With rounding, this solution is valid for the initial problem.

mylen [45]3 years ago

8 0

Answer:

x = 6, solution is not extraneous

Step-by-step explanation:

Given the equation 2 Square root of x minus 5 = 2

$2\sqrt{x-5} =2$ . solve for x

Divide by 2 on both sides

$\sqrt{x-5} =1$

Take square on both sides to remove square root

x-5= 1

add 5 on both sides

x= 6

Now we plug in x=6 in our original equation

$2\sqrt{6-5} =2$

2 = 2 is true

So x= 6 satisfies our given equation

X=6 is not an extraneous solution

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Solve the following inequality: 3x + 9 ≤7x -11.

m_a_m_a [10]

Answer:

x ≥ -3

Step-by-step explanation:

Step 1 :

Pulling out like terms :

1.1 Pull out like factors :

-4x - 12 = -4 • (x + 3)

Equation at the end of step 1 :

Step 2 :

2.1 Divide both sides by -4

Remember to flip the inequality sign:

Solve Basic Inequality :

2.2 Subtract 3 from both sides

x ≥ -3

Inequality Plot :

2.3 Inequality plot for

-4.000 X - 12.000 ≥ 0

6 0

3 years ago

Read 2 more answers

F(x)=( x^4 + 5x^3 - 2x^2 + 5x - 3) d(x)=( x^2 + 1) Divide f(x) by d(x)

VLD [36.1K]

f '(x) = [(-40x +11)(7x - 9) - 7(-4x +3)(5x + 1)]/(7x - 9)2
= [(-280x2 + 360x + 77x - 99) - 7(-20x2 - 4x + 15x + 3)]/(7x - 9)2
= [(-280x2 + 437x - 99) + (140x2 + 28x - 105x - 21)]/(7x - 9)2
= (-140x2 +360x - 120)/(7x - 9)2

i think thats how you would solve it
hope this helps tho:)

4 0

3 years ago

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,

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

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