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3 years ago

What is the formula to find the length of an arc when given the central angle and radius?

Mathematics

1 answer:

vazorg [7]3 years ago

3 0

$s = \dfrac{n}{360^\circ}2 \pi r$

where
s = arc length
n = measure of central angle
r = radius

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Write an equation of a line (in standard form) that has the same slope as the line 3x-5y=7 and the same y-intercept as the line

Vanyuwa [196]

Answer:

$3x-5y=-20$

Step-by-step explanation:

We can write the equation of a line in 3 different forms including slope intercept, point-slope, and standard depending on the information we have. We have two standard form equations which we will get a slope and a y-intercept from. We will convert each to slope intercept form to get the information. We will then write a new slope-intercept equation and convert to standard form.

3x-5y=7 has the same slope as the line. Let's convert.

$3x-5y=7\\3x-3x-5y=7-3x\\-5y=7-3x\\\frac{-5y}{-5}=\frac{7-3x}{-5} \\$

$y=\frac{3}{5}x -\frac{7}{5}$

The slope is $m=\frac{3}{5}$ .

2y-9x=8 has the same y-intercept as the line. Let's convert.

$2y-9x=8\\2y-9x+9x=8+9x\\2y=8+9x\\\frac{2y}{2}=\frac{8+9x}{2}$

$y=\frac{8}{2}+\frac{9x}{2} \\y=4+\frac{9}{2}x$

The y-intercept is 4.

We take $m=\frac{3}{5}$ and b=4 and substitute into y=mx+b.

$y=\frac{3}{5}x+4$

We now convert to standard form.

$-\frac{3}{5}x+y=\frac{3}{5}x-\frac{3}{5}x+4\\-\frac{3}{5}x+y=4$

For standard form we need the coefficients of x and y to be not zero or fractions. We need integers but the coefficient of x cannot be negative. So we multiply the entire equation by -5 to clear the denominators.

$-5(-\frac{3}{5}x+y=4)\\3x-5y=-20$

3 0

3 years ago

Solve the system of equations. −2x+5y =−35 7x+2y =25

Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

$\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}$

This system of equations can be solved in three different ways:

Graphing the equations (method used)
Substituting values into the equations
Eliminating variables from the equations

Graphing the Equations

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is $\text{y = mx + b}$ .

Equation 1 is $-2x+5y = -35$ . We need to isolate y.

$\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7$

Equation 1 is now $y=\frac{2}{5}x-7$ .

Equation 2 also needs y to be isolated.

$\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}$

Equation 2 is now $y=-\frac{7}{2}x+\frac{25}{2}$ .

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}$

$\bullet \ \text{For x = 0,}$

$\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5$

Now, we can place these values in our table.

As we can see in our table, the rate of decrease is $-\frac{2}{5}$ . In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract $-\frac{2}{5}$ from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be $y=-\frac{7}{2}x+\frac{25}{2}$ . Therefore, we just use the same process as before to solve for the values.

$\bullet \ \text{For x = 0,}$

$\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5$

And now, we place these values into the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1 Equation 2

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$ $\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

Therefore, using this data, we have one solution at (5, -5).

4 0

3 years ago

F(x) = 2x^2 – 5x – 3 g(x) = 2x^2 + 5x + 2 Find: (f/g) (x)

GrogVix [38]

Answer:

Since this means f of g so I will get the function g(x) and replace it with X in the function of f (x)

5 0

3 years ago

Devon worked for Mr.greene on four days last week.He was paid a total of $120 for the four days of work. If he worked 2 hours ea

sammy [17]

It'll be 60 if you divided there is your actually answer

8 0

3 years ago

Try This one out and I’ll give you brainliest no links or I will report you Thank you

sladkih [1.3K]

Answer:

32. steps below.

Step-by-step explanation:

So calculate 4(2x2^3-8)

4(2^4-8)

calculate

4(16-8)

Then get

4x8=32.

3 0

2 years ago

Read 2 more answers