Convert 4 5/6 to a fraction greater than one

What is the arc measure of \stackrel{\LARGE{\frown}}{AC}

algol [13]

Answer:

72°

Step-by-step explanation:

Given that the radius(r) of the circle is 10 units and the length of arc ABC is 16π

The length of arc ABC = $\frac{\theta}{360}*2\pi r$

Where θ is the central angle in degrees.

Since the length of arc ABC is 16π,

$16\pi=\frac{\theta}{360}*2\pi r\\16\pi=\frac{\theta}{360}*2\pi *10\\\theta=\frac{16\pi *360}{2 \pi *10} \\\theta=288^0$

The angle in a circle = 360°, therefore:

Central angle for arc AB (θ) = 360 - Central angle for arc ABC = 360 - 288 = 72°

Therefore the arc measure of arc AB is 72°

5 0

4 years ago

10 subtracted from the quotient of a number and -3 is at least -1

Mazyrski [523]

10 subtracted from the quotient of a number and -3 is at least -1 is:

$\left(\frac{x}{-3}\right)-10\ge \:-1\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-27\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-27]\end{bmatrix}$

Solution:

Given statement is:

10 subtracted from the quotient of a number and -3 is at least -1

Quotient means reuslt of dividing two numbers

Let the number be "x"

quotient of a number and -3 = $\frac{x}{-3}$

Therefore,

10 subtracted from $\frac{x}{-3}$ is at least -1

at least means "less than or equal to"

Which is translated into algebraic expression as:

$(\frac{x}{-3}) - 10\geq -1$

Solve the above inequality

$\left(\frac{x}{-3}\right)-10\ge \:-1\\\\\mathrm{Add\:}10\mathrm{\:to\:both\:sides}\\\\\frac{x}{-3}-10+10\ge \:-1+10\\\\\mathrm{Simplify}\\\\\frac{x}{-3}\ge \:9\\\\Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)$

When we multiply or divide by negative number. we have to flip the inequality sign

$\frac{x\left(-1\right)}{-3}\le \:9\left(-1\right)\\\\\mathrm{Simplify}\\\\\frac{x}{3}\le \:-9\\\\\mathrm{Multiply\:both\:sides\:by\:}3\\\\\frac{3x}{3}\le \:3\left(-9\right)\\\\\mathrm{Simplify}\\\\x\le \:-27$