What the value of w?

if the diameter is 20, the its radius must be half that or 10.

[tex]\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=5\pi \\ r=10 \end{cases}\implies \begin{array}{llll} 5\pi =\cfrac{\theta \pi (10)^2}{360}\implies 5\pi =\cfrac{5\pi \theta }{18} \\\\\\ \cfrac{5\pi }{5\pi }=\cfrac{\theta }{18}\implies 1=\cfrac{\theta }{18}\implies 18=\theta \end{array}[/tex]

Answer:

17 or 7

Step-by-step explanation:

if im right brainliest pls
?

Answer:

Given:

[tex] \\ {\tt{A = \left [ \: \begin{array}{ c c c} \: \: \: \: 2 & -3 & - 5 \\ - 1 & \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: 1& - 3 & - 4\end{array} \: \: \: \right] \: and \: B = \left[\begin{array}{c c c} \: \: \: \: 2 & - 2 & - 4 \\ - 1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & - 2 & - 3\end{array} \right]}}[/tex]

Matrix A is of order 3 × 3 and Matrix B is of order 3 × 3

To Show:

Formula Used:

[tex]\\ { \rm{ \left[ \begin{array}{c c c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array} \right] \times \left[ \begin{array}{c c c} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right] = \: \left[ \begin{array}{c c c} \: \: a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} & a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32} & a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33} \\ \: \: a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} & a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32}&a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33} \\ \: \: a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31}&a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32} & a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33}\end{array} \right]}}[/tex]

If A is a matrix of order a×b and B is a matrix of order c×d , then matrix AB exits and is of order a×d , if and only if b = c.

[tex] \: \: [/tex]

If A is a matrix of order a×b and B is a matrix of order c×d , then matrix BA exits and is of order c×b , if and only if d = a.

[tex] \: \: [/tex]

For Matrix AB, a = 3, b = c = 3, d = 3 , thus Matrix AB is of order 3×3

[tex]\\ { \rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right] \times \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] = \left[\begin{array}{c c c} \: \: \: \: 2(2)-3(-1)-5(1) & \: \: \: \: 2(-2)-3(3)-5(-2) & \: \: \: \: 2(-4)-3(4)-5(-3) \\ -1(2)+4(-1)+5(1) & -1(-2)+4(3)+5(-2) & -1(-4)+4(4)+5(-3)\\ \: \: \: \: 1(2)-3(-1)-4(1) & \: \: \: \: 1(-2)-3(3)-4(-2) & \: \: \: \: 1(-4)-3(4)-4(-3) \end{array} \right]}}[/tex]

[tex]\\ {\rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 4 + 3 - 5 & \: \: \: -4 -9 + 10 & \: \: -8 - 12 + 15 \\ -2 - 4 + 5 & \: \: \: \: \: \: + 2 + 12 - 10 & \: \: \: \: \: 4 + 16 - 15 \\ \: \: \: \: \: 2 + 3 - 4 & \: \: -2 - 9 + 8 & -4 - 12 + 12 \end{array} \right] = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right]}}[/tex]

[tex]\\ {\rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right] = Matrix \: A}}[/tex]

For Matrix BA, a = 3, b = c = 3, d = 3 , thus Matrix BA is of order 3 × 3

[tex]\\ {\rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] \times \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: \: 1 & -3 & -4 \end{array} \right]}}= \left[ \begin{array}{c c c} \: \: \: \: \: 2(2) -2(-1) -4(1) & \: \: \: \: 2(-3) -2(4) -4(-3) & \: \: \: \: 2(-5) -2(5) -4(-4) \\ \: -1(2) + 3(-1) + 4(1) & -1(-3) + 3(4) +4(-3) & -1(-5) + 3(5) + 4(-4) \\ \: \: \: \: \: 1(2) -2(-1) -3(1) & \: \: \: \: 1(-3) -2(4) -3(-3) & \: \: \: \: 1(-5) -2(5) -3(-4) \end{array} \right][/tex]

[tex]\\{ \rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 4 + 2 - 4 & \: \: \: -6 -8 + 12 & \: \: -10 - 10 + 16 \\ -2 - 3 + 4 & \: \: \: \: \: \: +3 + 12 - 12 & \: \: \: \: \: + 5 + 15 - 16 \\ \: \: \: \: \: 2 + 2 - 3 & \: -3 -8 +9 & \: \: \: \: \: -5 -10 + 12 \end{array} \right] = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right]}}[/tex]

[tex]\\{ \rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] = Matrix \: B}}[/tex]

Answer:

which one?

Step-by-step explanation:

Answer:

70

Step-by-step explanation:

60/40Ⅹ50=70

I hope this helps

Answer:

A. 9A² - 16B²

Step-by-step explanation:

(3A + 4B)(3A - 4B)

Distribute:

3A(3A - 4B)

4B(3A - 4B)

Multiply:

3A(3A - 4B)

3A(3A) + 3A(-4B)

9A² - 12AB

4B(3A - 4B)

4B(3A) + 4B(-4B)

12AB - 16B²

Combine like terms:

9A² - 12AB + 12AB - 16B² <== -12AB + 12AB cancels out

9A² - 16B²

Therefore, our final answer is 9A² - 16B²

Hope this helps!

Hey there!

We're given 2 points that the line intersects, so we can easily determine the line's slope.

This is the formula:

[tex]\bold{\displaystyle\frac{y2-y1}{x2-x1} }[/tex]

[tex]\bold{\displaystyle\frac{5-(-6)}{-4-(-9)}}[/tex]

[tex]\bold{\displaystyle\frac{5+6}{-4+9}}[/tex]

[tex]\bold{\displaystyle\frac{11}{5}}[/tex]

Hence, the answer is

[tex]\boxed{\boxed{\bold{\frac{11}{5} }}}[/tex]

Hope everything is clear.

Let me know if you have any other questions!

#LearningDoesn'tEnd

:-)

[tex]\rule{200}{2}[/tex]

Step-by-step explanation:

I hope its helped ! studied well.

Answer:

a: The tape diagram represents the story because there’s 5 boxes representing the 5 weeks and x-2 in each box since x represents the amount Andre’s neighbor will pay him weekly to mow the lawn. This all equals 40 because that’s how much money he has to save.

B: Andre’s neighbor pays him 10 dollars each week to mow the lawn.

Step-by-step explanation:

b: For this part you have to make an equation combining all the five weeks which would be 5x-10. Then you would make it equal 40 because that’s how much he needs to save 5x-10=40. The you solve to find x by adding 10 to both sides of the equation and get 5x = 50. Then if you divide 5 on both sides you get 10!
You can check you work by plugging in 10 as x into the diagram.

Answer:

Adjacent Angles are 2 angles with common sides and a common vertex that don't overlap in any way.

Step-by-step explanation:

Leave a comment if you need any more help. Give Brainiest, please.

Answer:

False

Step-by-step explanation:

2/10 is not equal to 4/16.

Answer:

Non-linear equations:

[tex]y=x^2 +2x-3[/tex]

[tex]3y=2^x[/tex]

Step-by-step explanation:

Linear equation: y = mx + b

Therefore, a linear equation has 2 variables that have no exponents.

So the non-linear equations are:

[tex]y=x^2 +2x-3[/tex]  (this is a quadratic equation - polynomial with degree 2)

[tex]3y=2^x[/tex] (this is an exponential equation)

Step-by-step explanation:

She decided to test this by 200 adults, so. Divide 6 by 500 = 0.012 (200)=2.4mins.

Answer: The 200 adults

Step-by-step explanation:

The volume of the cylinder is the amount of space on it

The volume of the oblique cylinder is 174.19 cubic units

The given parameters are:

AB = 16 units

Radius (r) = 2 units

[tex]\theta = 30^o[/tex]

Start  by calculating the height (h) using:

[tex]h = 16 * \cos(30^o)[/tex]

This gives

[tex]h = 13.86[/tex]

The volume of the cylinder is then calculated as:

[tex]V = \pi r^2 h[/tex]

This gives

[tex]V = 3.142 * 2^2 * 13.86[/tex]

[tex]V = 174.19248[/tex]

Approximate

[tex]V = 174.19[/tex]

Hence, the volume of the oblique cylinder is 174.19 cubic units

Read more about volumes at:

https://brainly.com/question/1972490

Answer:

D. 60  Mark brainliest<3

Step-by-step explanation:

Each neighbor buys 2 boxes which contains 6 candles meaning they in total contain 12.

5x12 = 60

Answer: 48x/64

Work (assuming ^ is multiplication):

12x^4/4x^16

12x ^ 4x = 48x

4 ^ 16 = 64

Answer:

maximum height is 45/4 in feet

The maximum height of the ball is 45/4 in feet

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

Given tha Pam kicked a soccer ball down the field. Part of the ball's path, including its highest point, is described by the function below, where f(x) is the height of the ball in feet.

F(x)= 12x^2+2x        

Thus for this quadratic , the max value will occur at x= -b/2a

=  - 2 / (2 x 1/12) = -12

Plug in -12 to find the max height

= (-12)^2 + 2 x-12

=  45/4 in feet

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Answer:

f(x)=x(x+4)(x+3)(x-4)(x-3)

Step-by-step explanation:

hello :

an  polynomial is : f(x)=(x+4)(x+3)(x-0)(x-4)(x-3)

means : f(x)=x(x+4)(x+3)(x-4)(x-3)

[tex]\text{If}~ \alpha ~\text{and}~ \beta~ \text{are the roots,}\\ \\x^2 -(\alpha + \beta)x +\alpha\beta = 0\\\\\text{For}~ x^2+3x-2=0\\\\\alpha +\beta=-\dfrac ba=-3\\ \\\alpha \beta = \dfrac ca = -2\\\\[/tex]

[tex]\text{So,~} \alpha^3+\beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 -\alpha \beta)\\ \\~~~~~~~~~~~~~~~~=(\alpha+\beta)\left[(\alpha +\beta)^2 -2\alpha \beta - \alpha \beta\right]\\\\~~~~~~~~~~~~~~~~=(\alpha+\beta)\left[(\alpha +\beta)^2 -3\alpha \beta \right]\\\\~~~~~~~~~~~~~~~~=-3\left[(-3)^2-3(-2)\right]\\\\~~~~~~~~~~~~~~~~=-3(15)\\\\~~~~~~~~~~~~~~~~=-45\\\\\text{And}~~~\alpha^3 \beta^3= (\alpha \beta )^3 = (-2)^3 = -8\\\\[/tex]

[tex]\text{Hence the equation whose roots are}~ \alpha^3 ~\text{and}~ \beta^3~ \text{is:}\\ \\ x^2-(\alpha^3 + \beta^3)x +\alpha^3 \beta^3 =0\\\\\implies x^2 -(-45)x+(-8)=0\\\\\implies x^2 +45x -8=0[/tex]

Answer:

The answer is C

Step-by-step explanation:

[tex]\frac{n}{20}*20 = 1 * 20[/tex]

1 x 20 = 20

Answer:

C) n = 20

Step-by-step explanation:

hope it helps!

Answer:

Step-by-step explanation:

Let the number = 'x'

Product of number and eight ⇒ 8*x = 8x

Nine more than the product of 8x ⇒add 9 to 8x

8x + 9 = 4

      8x = 4 - 9

     8x = -5

       x = -5/8

Answer:

Shorter leg = 20 cm.

Longer leg = 21cm.

Hypotenuse = 29 cm

Step-by-step explanation:

Let the shorter leg be x cm long

Then longer leg = x + 1 and hypotenuse = x + 9 cm long

So by Pythagoras:

x^2 + (x + 1)^2 = (x + 9)^2

0 = x^2 + 18x + 81 - x^2 - (x^2 + 2x + 1)

x^2 + 18x + 81 - x^2 - x^2 - 2x - 1 = 0

-x^2 + 16x + 80 = 0

x^2 - 16x - 80 = 0

(x - 20)(x + 4) = 0

So x = 20   (we ignore the negative root).

So longer leg = 20 + 1 and hypotenuse = 20 + 9.

I'll do the first three questions to get you started.

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

Question 1

x = horizontal component of the triangle

cos(angle) = adjacent/hypotenuse

cos(theta) = horizontal/hypotenuse

cos(40) = x/30

x = 30*cos(40)

x = 22.98133329 approximately

This is the horizontal force applies to the suitcase.

Work = force*displacement

Work =  22.98133329*120

Work = 2,757.7599948

Work = 2,758

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

Question 2

Add the drone's relative air velocity and the wind velocity vectors.

v+w = (13i + 3j) + (-7i - 2j)

v+w = (13i-7i) + (3j-2j)

v+w = 6i + j

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

Question 3

Check out the diagram below. Focus on the smaller triangle. Notice the placement of angle theta.

sin(theta) = opposite/hypotenuse

sin(40) = x/200

x = 200*sin(40)

x = 128.557521937308

x = 128.558

Answer:

g(x)=3x--2

Step-by-step explanation:

please mark me as brainlest

y - 6 = - 5 * x - 5 * (-3)

y - 6 = -5x + 15

y + 5x = 15 + 6

5x + y = 21

Answer:

5x + y = 21.

Step-by-step explanation:

y - 6 = -5(x - 3)

y - 6 = -5x + 15

5x + y - 6 = 15

5x + y = 15 + 6

5x + y = 21.

Answer:

120

Step-by-step explanation:

6p3 = 120

The number of the different combinations of the teams will be 26.

A permutation is an act of arranging objects or numbers in a particular order. Combinations are a method of selecting objects or numbers from a group of objects or collections without regard for their order.

Given that for a class activity, a teacher put students into teams of 5. The teams were made up of 2 students randomly chosen from group A and 3 students randomly chosen from group B. If group A has 4 students and group B has 6 students,

The number of possible combinations will be calculated as,

C = ⁶C₃ + ⁴C₂

C = ( 4 x 3 ) / 2 + ( 6x 5 x 4 ) /( 3x 2 )

C = 6 + 20

C = 26

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Answer:

First one: 27

Second one: 1/8

Step-by-step explanation:

Hope this helps! :)