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3 years ago

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: