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hram777 [196]
3 years ago
10

A manager needs to rope off a rectangular section for a private party. The length of the section must be 7.6 meters. The manager

can use no more than 28 meters of rope. What inequality could you use to find the possible width, w, of the roped-off section?
Mathematics
1 answer:
Andrew [12]3 years ago
5 0

Answer:

W\leq 6.4\ m

Step-by-step explanation:

we know that

The perimeter of a rectangle is equal to

P=2(L+W)

we know that

L=7.6\ m

P\leq28\ m

substitute

2(7.6+W)\leq28

solve for W

(7.6+W)\leq14

W\leq14-7.6

W\leq 6.4\ m

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Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3
nlexa [21]

Answer:

(x,y) = (5,-6)

Step-by-step explanation:

\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\

                     ~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)

\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\

\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5

y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6

\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)

7 0
2 years ago
What would I have to do here? Somebody please help me! (Look at the bottom problem not the top)
Zinaida [17]
Alright, let's do this! If $25 is one fifth of how much the game system cost then, multiply 25 times 5. And you should get your answer. 
I hope this helps. 
8 0
3 years ago
What is 67,234.63992 rounded to the nearest thousandth
GuDViN [60]
67,234.63992 to the nearest thousandth is 67,234.64.
7 0
3 years ago
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PLEASE HELP ME I NEED THIS ANSWER ASAP!!! What is the area of this figure?
AnnZ [28]
A i am sure i hope this helps

8 0
3 years ago
Hayden will attend a craft show the cost of admission ticket is eight dollars the cost of a raffle ticket for a handmade quilt a
ruslelena [56]

Answer: 7 tickets

Step-by-step explanation:

First you need to remove the cost of entrance from the $45 that Hayden has:

= 45 - 8

= $37

The $37 is the amount that is left for Hayden to be able to buy tickets for the raffle.

The tickets cost $5 each so the highest number of tickets that can be bought with $37 is:

= 37 / 5

= 7 tickets

6 0
3 years ago
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