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3 years ago

Can someone help me solve this please

Mathematics

1 answer:

Soloha48 [4]3 years ago

8 0

Answer:

Step-by-step explanation:

We khow that :

A= (θ/360)*π*r² where A is the area of the sector and θ is the measure of the angle

A=(120/360)*π*10.5²= 115.45≈115.5

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This is LCM and GCF I think

Ugo [173]

Yes it is LCM and GCF because you are trying to a factor that they both have

3 0

3 years ago

Read 2 more answers

If you good with 7th grade math pls help

DochEvi [55]

289 196 248 379 319 276 198 349

(in order) 196, 198, 248, 276, 289, 319, 349, 379 there are eight numbers.

--------------------------------------------------------------------------------------

196, 198, 248, 276

since the median is between 198 and 248, we must add both numbers together and divide them by two:

(198 + 248) ÷ 2

= 446/2

<h2>= 223 (first quartile)</h2>

--------------------------------------------------------------------------------------

289, 319, 349, 379

since the median is between 319 and 349, we must add both numbers together and divide them by two:

(319 + 349) ÷ 2

= 668/2

<h2>= 334 (third quartile)</h2>

--------------------------------------------------------------------------------------

Interquartile range:

I double-checked my answers btw

3 0

3 years ago

If anyone knows this I will give them brainiest

statuscvo [17]

Answer:

x = 12

m(QS) = 52°

m(PD) = 152°

Step-by-step explanation:

Recall: Angle formed by two secants outside a circle = ½(the difference of the intercepted arcs)

Thus:

m<R = ½[m(PD) - m(QS)]

50° = ½[(12x + 8) - (4x + 4)] => substitution

Solve for x

Multiply both sides by 2

2*50 = (12x + 8) - (4x + 4)

100 = (12x + 8) - (4x + 4)

100 = 12x + 8 - 4x - 4 (distributive property)

Add like terms

100 = 8x + 4

100 - 4 = 8x

96 = 8x

96/8 = x

12 = x

x = 12

✔️m(QS) = 4x + 4 = 4(12) + 4 = 52°

✔️m(PD) = 12x + 8 = 12(12) + 8 = 152°

3 0

3 years ago

Here is 3 & 4 I NEED DONE ASAP!

Marrrta [24]

Given matrices are from here

brainly.com/question/18267865

=====================================================

Problem 3

B - C = DNE

We cannot subtract matrices of different sizes. Matrix B is 2x2 while C is 3x2. Both matrices must have the same number of rows, and they must also have the same number of columns. The matrices don't have to be square.

------------------------------

$A+B = \begin{bmatrix}-5 & 6\\7 & 4\end{bmatrix}$

You add the corresponding elements. For instance, in the top left corner we have -1+(-4) = -5. The other entries are treated in a similar manner.

------------------------------

$-2E = \begin{bmatrix}6 & -4 & 8\\-12 & 14 & -16\\-10 & -18 & 20\end{bmatrix}$

You get this from multiplying each entry in matrix E by -2. Eg: top left corner has -2*(-3) = 6

------------------------------

$CD = \begin{bmatrix}4 & 7 & 8\\ 1 & 3 & 4\\14 & 17 & 16\end{bmatrix}$

Matrix C has 3 rows and D has 3 columns. The final answer will be size 3x3

To generate each value in the answer matrix, you'll highlight rows of C to pair with columns of D. Then you'll multiply out the corresponding values, after which you add those products. This is done for every entry in the answer shown above.

For example, the first row of C is highlighted and the second column of D is highlighted. Those values pair up and multiply getting -1*(-1) + 2*3 = 1+6 = 7, which goes in the first row and second column of the answer matrix. The other entries are handled in a similar fashion.

------------------------------

$\det(B) = -46$

The 2x2 matrix determinant formula is $\begin{vmatrix}a & b\\c & d\end{vmatrix} = a*d - b*c$

In this case, a = -4, b = 6, c = 5, d = 4.

------------------------------

$B^{-1} = \begin{bmatrix}-2/23 & 3/23\\ 5/46 & 2/23\end{bmatrix}$

Swap the top left and bottom right corners of matrix B. Change the sign of the other two corner values. Then multiply each entry by 1/d where d is the determinant found back in part (e) above. Be sure to reduce any fraction as much as possible.

=====================================================

Problem 4

Answers: x = 2 and y = -10

-----------------

Work Shown:

$\begin{bmatrix}2x & 4\\-5 & -2\end{bmatrix}+\begin{bmatrix}3 & -7\\11 & y-1\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\\begin{bmatrix}2x+3 & 4+(-7)\\-5+11 & -2+(y-1)\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\\begin{bmatrix}2x+3 & -3\\6 & y-3\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\$

In the top left corners of each matrix, in line 3, we have 2x+3 = 7 which solves to...

2x+3 = 7

2x = 7-3

2x = 4

x = 4/2

x = 2

In the bottom right corners, we have y-3 = -13 which solves to....

y-3 = -13

y = -13+3

y = -10

4 0

4 years ago

Drag each number to the correct location on the equations. Each number can be used more than once, but not all numbers will be u

VARVARA [1.3K]

Answer:

a=$10

c=$3

Step-by-step explanation:

Step one:

given data

let adults tickets be a

and child tickets be c

Roy's purchase

6a+2c=66------1

Elisa's purchase

5a+4c=62-------2

Hence, the system of equations

6a+2c=66------1

5a+4c=62-------2

solving simultaneously

6a+2c=66------1 x2

5a+4c=62-------2 x1

12a+4c=132

-5a+4c=62

7a=70

divide both sides by 7

a=$10

put a=70 in 1

6(10)+2c=66------1

60+2c=66

2c=66-60

2c=6

c=$3

5 0

3 years ago

Read 2 more answers