Enter the equation of the line in slope-intercept form.

xxTIMURxx [149]

The pythagorean theorem states that a^2+b^2=c^2.
3^2 is 9.
9+9=18
The square root of 18 is approximately 4.24.
a. $\sqrt{18}$
b. $3 \sqrt{2}$
c.4.24

6 0

4 years ago

Calculate area and perimeter

mina [271]

Answer:

area ≈ 12.505

perimeter ≈ 16.1684

Step-by-step explanation:

We are given

- the radius of the circle (and therefore area of the circle)

- the area of the triangle

We want to find

- angle AOB/AOT. We want to find this because 360/the angle gives us how many OABs fit into the circle. For example, if AOT was 30 degrees, 360/30 = 12 (there are 360 degrees in a circle, so that's where 360 comes from). The area of the circle is equal to πr² = π6² = 36π, and because AOT is 30 degrees, there are 12 equal parts of sector OAB in the circle, so 36π/12=3π would be the area of the sector. A similar conclusion can be reached from the circumference instead of the area to find the distance between A and B along the circle, and OA + AB + BO = the perimeter of the minor sector.

First, we can say that OAT is a right triangle because a tangent line is perpendicular to the line from the center to the point on the circle, so AT is perpendicular to OA. This forms two right angles, one of which is OAT

One thing that we can start to solve is AT. We know that the area of a triangle is equal to base * height /2, and the height of this triangle is AO, with the base being AT. Therefore, we can say

15 = AO * AT / 2

15 = 6 * AT / 2

15 = 3 * AT

divide both sides by 3 to isolate AT

AT = 5

Because OAT is a right triangle, we can say that the hypotenuse ² = the sum of the squares of the two other lengths. The hypotenuse is opposite of the largest angle (in this case, the right angle, as in a right triangle, the right angle is always the largest), so it is OT in this case. The other two sides are OA and AT, so we can say that

OA² + AT² = OT²

5²+6² = OT²

25+36=61=OT²

square root both sides

OT = √61

Next, the Law of Sines states that

sinA/a = sinB/b = sinC/c with angles A, B, and C with sides a, b, and c. Corresponding sides are opposite their corresponding angles, so in this case, AT corresponds to angle AOT, OT corresponds to angle OAT, and AO corresponds to angle ATO.

We want to find angle AOT, as stated earlier, so we have

sin(OAT)/OT = sin(ATO)/OA = sin(AOT)/AT

We know the side lengths as well as OAT/sin(OAT) and want to figure out AOT/sin(AOT), so one equation that helps us get there is

sin(OAT)/OT = sin(AOT)/AT, encompassing our 3 known values and isolating the one unknown. We thus have

sin(90)/√61 = sin(AOT) /5

plug in sin(90) = 1

1/√61 = sin(AOT)/5

multiply both sides by 5 to isolate sin(AOT)

5/√61 = sin(AOT)

we can thus say that

arcsin(5/√61) = AOT ≈39.80557

As stated previously, given ∠AOT, we can find the area and perimeter of the sector. There are 360/39.80557 ≈ 9.04396 equal parts of sector OAB in the circle. The area of the circle is πr² = 36π, so 36π / 9.04396 ≈ 12.505 as the area. The circumference is equal to π * diameter = π * 2 * radius = 12 * π, and there are 9.04396 equal parts of arc AB in the circumference, so the length of arc is 12π / 9.04396 ≈ 4.1684. Add that to OA and OB (both are equal to the radius of 6, as any point from the center to a point on the circle is equal to the radius) to get 6+6 + 4.1684 = 16.1684 as the perimeter of the sector

5 0

3 years ago

Directions: Find the value of x.

Umnica [9.8K]

I think it would be 12.207… use the Pythagorean theorem (a^2+b^2=c^2) so 10^2+7^2
100+49=149
Then take square root of 149 to undo the the squaring of c and it should be around 12.207 :)

4 0

2 years ago

Read 2 more answers

3 determine the highest real root of f (x) = x3− 6x2 + 11x − 6.1: (a) graphically. (b) using the newton-raphson method (three it

Juliette [100K]

(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)

(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...

... x ≈ 3.0473167

(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...

... x ≈ 3.2291234

(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...

... x ≈ 3.0477377

(e) Using <em>Mathematica</em>, the roots are found to be as shown in the second attachment. The highest root is about ...

... x ≈ 3.0466805180

_____

<em>Comment on these methods</em>

Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.

The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.

We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).

6 0

3 years ago

Help me please I all ready submit all my question on my profile go see it it is due right now I will mark you as a brainlest rig

pashok25 [27]

Answer:

ok

Step-by-step explanation:

6 0

3 years ago