Find the value of X. Round your answer to the nearest tenth please

Hey there!

Answer: = 15.3296% decrease

How to Calculate Percentage Decrease

=> 539.22 - 456.56

=> 79.66

=> 79.66 ÷ 539.22

=> 0.153296

=> 0.153296 × 100

=> 15.3296%

If the percentage is negative, it means there was an increase and not an decrease.

Therefore, percentage decrease is 15.3%

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:

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)

Based on the outside angles theorem, the value of x is: 42°

The outside angles theorem states that the measure of the angle that is formed outside a circle when two tangents or secants intersect is half the measure of the diffrence of the intercepted arcs.

Given the following:

Therefore, based on the outside angles theorem, we would have the following:

x = 1/2(222 - 138)

x = 42°

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Answer: there is something off screen i cant see so i cant help you 183- what?

Step-by-step explanation:

Answer:

19 and 13

Step-by-step explanation:

width = x

length = x + 6

2x + 2(x + 6) = 64

2x + 2x + 12 = 64

4x = 64 - 12

4x = 52

x = 13

x + 6 = 19

Answer:

19 and 13

Step-by-step explanation:

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.

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

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:

the answer b

Step-by-step explanation:

te answer b because I guessed

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:

Answer:

which one?

Step-by-step explanation:

Answer:

False

Step-by-step explanation:

2/10 is not equal to 4/16.

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:

Can you use a specific question please

Step-by-step explanation:

Answer: = 17 7/

Step-by-step explanation:

8 1/5 + 9 1/2 = 177/

10

= 17 7/

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:

un rhgdb tennis felicidades ID remix estoy do do ID do ID Rockband do ID do TX

Answer:

median is 20

Step-by-step explanation:

due to both of them having it in there 1 time

Answer:

The Median Age: Yoga Class (31)      Dance Class (30)

The Range: Yoga Class (17-44)        Dance Class (20-47)

Step-by-step explanation:

Answer:

56

Step-by-step explanation:

Step 1: Order the data

43,49,55,58,64,67

Step 2: Count all numbers and add them up

43+49+55+58+64+67=336 and we have 6 numbers in total so... 336 ÷ 6

Step 3: Divide!

336 ÷ 6 = 56

Hope this helps you!

he was never late negative to affirmative

Answer:

an+1 = 4an

Step-by-step explanation:

an+1 = 4an        commonratio:   r=4

2

2x4=8

8x4=32

32x4=128

Answer:

If you buy 3 pounds for $2, and have 8 dollars to spend, then multiply 2x4 to get $8. Since we multiplied by 4 for the money, we do the same thing for the pounds of grapes. So, we multiply 3x4 to get 12 pounds. So, you can buy 12 pounds of grapes for $8.

Step-by-step explanation:

Hope this helps! :)

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Answer:

Step-by-step explanation:

75-52.5=x

The probability that the next customer will pay with a debit card or a credit card is 12%.

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

The probability that the next customer would pay with a debit card or a credit card = (number of customers that paid with a debit card + number of customers that paid with a credit card) / total number of customers

(8 + 2)/ 83 = 0.12 = 12%

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A. The amount of money that Betsy should invest n B-rated bonds at 13% is $20,000.

B. The amount of money that Betsy should invest in a certificate of deposit (CD) at 5% is $50,000.

Certificates of deposit (CDs) are short-term investments in financial institutions (money market) while Bonds are long-term investments in the stock exchange (capital market).

CDs pay reduced interest rates when compared to bonds'.

The amount required per year in interest = $5,000

Total investible funds = $70,000

Interest rate of B-rated bonds = 13%

Interest rate in CD = 5%

Since Betsy is not interested in realizing more than $5,000 in interest per year, she should invest $50,000 in the CD and $20,000 in the B-rated bonds.

The interest in CD = $2,500 ($50,000 x 5%)

The interest in bonds = $2,600 ($20,000 x 13%)

Total interest in CD and bonds = $5,100 ($2,500 + $2,600)

Thus, Betsy should invest $50,000 in the CD and $20,000 in the B-rated bonds.

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Answer:

See below ↓

Step-by-step explanation:

3.

4.

5.

Answer: 48x/64

Work (assuming ^ is multiplication):

12x^4/4x^16

12x ^ 4x = 48x

4 ^ 16 = 64

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]