The circumference formula for a circle is C=πd where C is the circumference, d is the diameter of the circle and π is a constant. If you plug in 6 for C and solve the equation for d like:

6= πd and then divide both sides of the equation by π you get that d = 1.90

To find the central angle of an arc you would use the equation S = rθ where S is the length of the arc, r is the radius of the circle, and θ is the measure of the angle which in this case is unknown. So with S = 1 and r = d/2 = 1.90/2 = 0.9549 you would have an equation that looks like this:

1 = 0.9549θ.

If you divide both sides of the equation by 0.9549 you get θ = 1.047197 radians.

The question asked for the angle measure in degrees so you would need to convert the angle measure to degrees by multiplying the degree measurement by 180/π

1.047197 x 180/π = 60°

8 0

2 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

Basile [38]

1.Disc method.
In this method the volume is given by:

$\boxed{V=\pi\int\limits_a^b\big[f(x)\big]^2}$

so:

$V=\pi\int\limits_1^3x^4\,dx=\boxed{\pi\int\limits_1^3\big[x^2\big]^2\,dx}$

A) Function $f(x)=x^2$ over the interval $[1,3]$
B) We use disk method and f(x) is function of variable x, so we rotate the curve about the x-axis.

2. Shell method.

In this case volume is given by:

 $\boxed{V=2\pi\int\limits_a^bx\cdot f(x)\,dx}$

So there will be:

$V=\pi\int\limits_1^3x^4\,dx=\dfrac{2}{2}\cdot\pi\int\limits_1^3x^4\,dx=2\pi\int\limits_1^3\dfrac{x^4}{2}\,dx=
\boxed{2\pi\int\limits_1^3x\cdot\dfrac{x^3}{2}\,dx}$

A) Function $f(x)=\dfrac{x^3}{2}$ over the interval $[1,3]$
B) We use shell method and f(x) is function of variable x, so we rotate the curve about the y-axis.

3 0

2 years ago

Evaluate the cosine if the angle of rotation which contains the point (9, -3) on its terminal side

Liono4ka [1.6K]

so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant

$\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{9}}{\stackrel{hypotenuse}{\sqrt{90}}}\implies \stackrel{\textit{rationalizing the denominator}}{\cfrac{9}{\sqrt{90}}\cdot \cfrac{\sqrt{90}}{\sqrt{90}}\implies \cfrac{9\sqrt{90}}{90}}\implies \cfrac{\sqrt{90}}{10}\implies \cfrac{3\sqrt{10}}{10}$

6 0

2 years ago