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2 years ago

8

Jamal's Lawn Service often fertilizes lawn in the spring. Jamal charges $3.50 per square yard. How much will Jamal charge to fer

tilize the yard pictured below

2 answers:

wariber [46]2 years ago

8 0

24.5in^2 because you do length times width which is 4*3

Lisa [10]2 years ago

5 0

28$ (I have to type more than 20 letters for a awnser sorry for this extra part)

You might be interested in

Plss help i’ll give brainliest thank you!

bija089 [108]

Answer:

36

Step-by-step explanation:

The formula of LCM is LCM(a,b) = ( a × b) / GCF (a,b)

The Greatest Common Factor of (9,12) = 3

Therefore

LCM of 9 and 12 = (9 × 12) / 3

= 108/3 = 36

LCM of 9 and 12 = 36

Irrespective of the method, the solution to our question LCM of 9 and 12 is the same.

Therefore, LCM of 9 and 12 is 36

5 0

2 years ago

Britney is solving a quadratic equation. Her first

Liula [17]

Britney used commutative property of addition and distributive property of multiplication over addition to get to step

The answer is II and IV ⇒ answer 4

Step-by-step explanation:

The quadratic equation is:

3x² - 8 - 10x = 3(2x + 3)

Britney's step is:

3x² - 10 x - 8 = 6x + 9

Lets explain the meaning of:

1. Commutative property of addition is: you can add a and b or b and a

both give the same answer

2. Distributive property of multiplication over addition is: when the

number a is multiplied by the sum of two numbers (b + c), the

first number a can be distributed to both b and c and multiplied

by each of them separately, then adding the two products ab and

ac together for the same result as multiplying the first number a by

the sum (b + c) ⇒ a(b + c) = ab + ac

3. Addition property of equality is: the property that states if you

add the same number to both sides of an equation, the sides

remain equal

4. Multiplication property of equality is: The property that states if

you multiply both sides of an equation by the same number, the

sides remain equal

Now let us explain Britney's step

∵ She change the places of the terms - 10x and - 8

- She arranges the terms of the left hand side from greatest power

of x to the numerical term

∴ She used the commutative property of addition

∵ She multiplied 3 by 2x and multiplied 3 by 3

- She multiplied the number out side the bracket by each of the

two terms in the bracket

∴ She used the distributive property of multiplication over addition

Britney used commutative property of addition and distributive property of multiplication over addition to get to step

The answer is number 4) II and IV

Learn more:

You can learn more about quadratic equation in brainly.com/question/8196933

#Learn withBrainly

8 0

3 years ago

Read 2 more answers

Liz was testing Jane on solving with equations. She gave Jane the

GuDViN [60]

Answer:

x/7+13=32

Step-by-step explanation:

work backward so subtract 13 from 32 and get 19 multiply it by 7 and get 133, to make sure your answers right just substitute 133 back into the equation

7 0

2 years ago

What is the solution to the system of equations below?<br><br> x = 5y - 2 AND 3x - 4y = 27 *

Deffense [45]

Answer:

x = 13, y = 3

Step-by-step explanation:

3(5y - 2) - 4y = 27

15y - 6 - 4y = 27

11y = 33

y = 3

x = 15 - 2

x = 13

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5C%5C%20%7B%20%5Crm%7BIf%20%20%5C%3A%20%7D%7B%5Ctt%7BA%20%3D%20%5Cleft%20%5B%20%5C%3A%20%5

RUDIKE [14]

Answer:

Given:

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

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

To Show:

Matrix AB = A , BA = B

Formula Used:

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

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.

$\: \:$

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.

$\: \:$

<h2>____________________________________________________________________</h2>

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

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

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

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

<h3>Matrix AB = Matrix A</h3><h2>____________________________________________________________________</h2>

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

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

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

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

<h3>Matrix BA = Matrix B</h3><h2>____________________________________________________________________</h2>

3 0

2 years ago