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neonofarm [45]
3 years ago
9

Find an exact solution for √5-1/x = √5/2 Then find the approximate solution to the nearest tenth.

Mathematics
1 answer:
viktelen [127]3 years ago
3 0

Step-by-step explanation:

\frac{ \sqrt{5 - 1 } }{x}  =  \frac{ \sqrt{5} }{2}  \\  \\ x \sqrt{5 }  = 2 \sqrt{5}  - 2 \\ x = 2 -  \frac{2 \sqrt{5} }{5}

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9(4b-1)=2(b+3) Slove for b plz answer as quick as u can please
Jobisdone [24]

36b - 9 = 2b + 6 \\ 34b = 15 \\ b =  \frac{15}{34}
3 0
3 years ago
Find the area of an equilateral triangle with radius 22 cm. Round to the nearest whole number.
earnstyle [38]

Answer:this item, we are given with the radius equal to 22 cm.

In this measurement of the radius is the hypotenuse of the 30°-60°-90° triangle formed with the half the measurement of the side of the equilateral triangle being opposite to the 60°  and equal to the hypotenuse times (1/2)(√3)

Step-by-step explanation:

If we let s be the side of the triangle then,

                 s/2 = r(1/2)(√3)

Multiplying the equation by 2,

                 s = r√3

Substituting,

              s = (22 cm)(√3) = 22√3

The area of the equilateral triangle is computed through the equation,

         A = (√3 / 4)(s²)

Substituting,

         A = (√3 / 4)(22√3)² = 628.7 cm²

Therefore, the answer to this item is the first choice. 

4 0
2 years ago
Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=x^2+y

\dfrac{\partial f}{\partial y}=y^2+x

\dfrac{\partial f}{\partial z}=ze^z

Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

So the scalar potential function is

\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
2 years ago
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Keith_Richards [23]

Answer:

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Step-by-step explanation:

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3 years ago
Find the weekly wage before deductions for following data.
andrew11 [14]

Answer:

Did I do something wrong because you never did text me back

Step-by-step explanation:

$34

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6 0
3 years ago
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