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2 years ago

In circle K with m JKL = 112 and JK = 13 units find area of sector JKL.

Mathematics

1 answer:

svetoff [14.1K]2 years ago

6 0

Answer:

I65.18

Step-by-step explanation:

I found this online I hope this helps

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(b) For those values of k, verify that every member of the family of functions y = A sin(kt) + B cos(kt) is also a solution. y =

frutty [35]

Answer:

Check attachment for complete question

Step-by-step explanation:

Given that,

y=Coskt

We are looking for value of k, that satisfies 4y''=-25y

Let find y' and y''

y=Coskt

y'=-kSinkt

y''=-k²Coskt

Then, applying this 4y'"=-25y

4(-k²Coskt)=-25Coskt

-4k²Coskt=-25Coskt

Divide through by Coskt and we assume Coskt is not equal to zero

-4k²=-25

k²=-25/-4

k²=25/4

Then, k=√(25/4)

k= ± 5/2

b. Let assume we want to use this

y=ASinkt+BCoskt

Since k= ± 5/2

y=A•Sin(±5/2t)+ B •Cos(±5/2t)

y'=±5/2ACos(±5/2t)-±5/2BSin(±5/2t)

y''=-25/4ASin(±5/2t)-25/4BCos(±5/2t

Then, inserting this to our equation given to check if it a solution to y=ASinkt+BCoskt

4y''=-25y

For 4y''

4(-25/4ASin(±5/2t)-25/4BCos(±5/2t))

-25A•Sin(±5/2t)-25B•Cos(±5/2t).

Then,

-25y

-25(A•Sin(±5/2t)+ B •Cos(±5/2t))

-25A•Sin(±5/2t) - 25B •Cos(±5/2t)

Then, we notice that, 4y'' is equal to -25y, then we can say that y=Coskt is a solution to y=ASinkt+BCoskt

4 0

2 years ago

What is the surface area of the net? *inches squared*

serious [3.7K]

9514 1404 393

Answer:

75 in^2

Step-by-step explanation:

The central vertical rectangle (including "ears") has dimensions

3 in wide by (9+2+2) = 13 in tall

Its area is

A = LW = (3 in)(13 in) = 39 in^2

The two rectangles either side of that have dimensions 2 in by 9 in. The area of each of them is

A = LW = (2 in)(9 in) = 18 in^2

The total net area is the sum of the areas of the parts:

left rectangle + central rectangle + right rectangle

= 18 in^2 + 39 in^2 + 18 in^2 = 75 in^2 . . . . surface area of the net

3 0

2 years ago

Show me how to solve 10>x/2 +4

postnew [5]

Treat this as if the inequality sign was the same thing as an equal sign. Most of the equality rules apply to inequalities (with a few exceptions)

$10>\dfrac{x}{2}+4$

Subtract both sides by 4

$10-4>\dfrac{x}{2}+4-4$

$6>\dfrac{x}{2}$

Multiply both sides by 2

$2\times6>\dfrac{x}{2}\times2$

$12>x$

We want the unknown variable on the left side (because it looks nicer)

$x$

5 0

3 years ago

Read 2 more answers

In 4 games, Larry averaged 23 PPG (points per game). How many points does he need to score in the last game to have an average o

ivann1987 [24]

Larry needs to score 33 points in the last game to have an average of 25 PPG.

To find the answer, we can solve the equation:

$\frac{(4)23+p}{5}=25$

(where p is how many points he needs to score in the last game)

To find the mean of a group of numbers, you add them all up and divide the total by the number of numbers there are. Since Larry averaged 23 PPG in 4 games, we can multiply 23 by 4 to get the total of the first 4 games from the data. Then, we find p, which we will add to get our final total. Then, you divide by the 5 games.

$\frac{(4)23+p}{5}=25$
$\frac{92+p}{5}=25$
${92+p}=125$
$p=33$

First, I simplified 4 x 23 to get 92. Then, I multiplied each side by 5 to get rid of the denominator. Finally, I subtracted 92 from each side to isolate p, and found that p = 33.

8 0

3 years ago

Read 2 more answers

If you graph the system of inequalities given below, which points would lie in the solution set? Select all the correct answers.

vivado [14]

Its just a matter of subbing...but keep in mind, for it to be a solution, it has to satisfy all 3 inequalities

x - y > = -4
2x - y < = 5
2y + x > 1

solutions are : (-1,3) and (3,5)

5 0

2 years ago

Read 2 more answers