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3 years ago

A rectangular garden has a width that is 8 feet less than twice the length. Find the dimensions if the perimeter is 20 feet.

Mathematics

1 answer:

krok68 [10]3 years ago

4 0

Answer:

length = 6 feet

width = 4 feet

Step-by-step explanation:

To find the dimensions of the rectangle, you can set up a variable and an expression that will equal the perimeter of the figure. Given l = length, width is '8 feet less than twice the length' or w = 2l - 8.

The general formula for perimeter of a rectangle is:

2w + 2l = P, where w = width and l = length

Using the variable and expression from above and the perimeter of 20 feet:

2(2l - 8)+ 2l = 20

distribute: 4l - 16 + 2l = 20

combine like terms: 6l - 16 = 20

add 16 to both sides: 6l - 16 + 16 = 20 + 16 or 6l = 36

divide and solve for 'l': 6l/6 = 36/6, or l = 6 feet

solve for 'w': w = 2(6) - 8 or w = 12 - 8 = 4 feet

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A mother decides to teach her son about a letter each day of the week. She will choose a letter from the name of the day. For ex

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Answer:

Check Explanation

Step-by-step explanation:

The days of the week include

Sunday

Monday

Tuesday

Wednesday

Thursday

Friday

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a) So, the possible letters to be taught include

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All the letters that cannot be taught using this method because no day of the week contains them include

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Hope this Helps!!!

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3 years ago

1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6

meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

x= -3

Putting value of x in equation 1

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

Let

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ = A $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = X and $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$ = B

Then AX= B

or X= A⁻¹ B

where A⁻¹= adj A/ ║A║ where mod A≠ 0

adj A= $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$ $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}42\\0\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-3\\0\\\end{array}\right]$

From here x= -3 and y= 0

Solution Set = [(-3,0)]

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