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3 years ago

A sector has an area of 1/2 π and the central angles of 1/9 π radians. What is the area of the circle?

Mathematics

1 answer:

zhenek [66]3 years ago

6 0

Answer:

Area of circle is 9π

Step-by-step explanation:

The area of sector of a circle is given by $A=\frac{1}{2}r^2\theta$

Plugging the given values in the formula, we get

$\frac{1}{2}\pi=\frac{1}{2}r^2\cdot\frac{1}{9}\pi$

Cancel out 1/2 and π from both sides, we are left with

$1=r^2\cdot\frac{1}{9}$

Cross multiplying, we get

$r^2=9\\\\r=3$

Therefore, area of circle is given by

$A=\pi r^2\\\\A=\pi(3)^2\\\\A=9\pi$

Area of circle is 9π

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3 years ago

Read 2 more answers

Can someone help me please? Use the linear combination method to solve the system of equations. Pretty please show your work and

Pavel [41]

Answer:

(x, y) = (4, -3)

Step-by-step explanation:

x-3y=13
2x+4y=-4

First of all, look at the given equations. Here, we see that the second equation has coefficients that all have a factor of 2. If we divide that out, we get an equation that has an x-coefficient of 1, matching the x-coefficient in the first equation.

Here is the reduced second equation:

x +2y = -2

Now, if we subtract one equation from the other, the variable x will be eliminated. We want to choose that subtraction wisely.

We note that the y-coefficient in the first equation is less than that in the second equation. If we subtract the first equation from the second, the result will have a positive y-coefficient:

(x +2y) -(x -3y) = (-2) -(13)

5y = -15 . . . . . . . . . . . . . . . . simplify. Note that the x-variable is eliminated, which is the purpose of this exercise.

y = -3 . . . . . . . divide by 5

We can use this value in the reduced second equation to find the value of x:

x +2(-3) = -2

x = 4 . . . . . . . . . add 6 and simplify

The solution is (x, y) = (4, -3).

_____

I find a graphing calculator provides an easy and reliable check of the answer.

_____

Comment on linear combination

This method is often called "elimination," because the purpose of combining the equations in a particular way is to eliminate one of the variables. This requires you look at the coefficients of the variables and devise a plan to combine them so the resulting coefficient for one of the variables is zero.

In the worst case, you can combine ...

ax +by = c
dx +ey = f

by multiplying the second equation by a and the first by -d:

a(dx +ey) -d(ax +by) = a( f) -d(c)

y(ae -bd) = fa -cd . . . . . . simplify; x is eliminated

This sort of approach results in a formula for the solution known as Cramer's Rule.

y = (fa-cd)/(ae-bd)

The corresponding solution for x is ...

x = (ce-bf)/(ae-bd)

The point of looking at the equations first is that you can often choose which variable to eliminate and what multiplier to use to minimize the amount of arithmetic involved—as we did above.