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Bogdan [553]
3 years ago
6

A rectangle has a length of 11 feet and a perimeter of 48 feet.What is the width in feet of the rectangle?

Mathematics
1 answer:
Art [367]3 years ago
7 0
The width is 37
if the length is 11
and the perimeter is 48 and u need to find the width what u do is
p=48
L=11
48-11=37
so the formula is
2L + 2w=p
11 + 37= 48
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A restaurant catered a party for 40 people. A child’s dinner (c) cost $11 and an adult’s dinner (a) cost $20. The total cost of
11111nata11111 [884]

a+c=40

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440-11a +20a=728

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Tính : 2020^3-1/2020+2021
pochemuha

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2020³-1/4041

Step-by-step explanation:

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In 3-4 sentences, describe how you would find a line parallel to the line 2x + 5y = 15 that goes through the point (-10, 1). Be
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the parallel line is 2x+5y+15=0.

Step-by-step explanation:

ok I hope it will work

soo,

Solution

given,

given parallel line 2x+5y=15

which goes through the point (-10,1)

now,

let 2x+5y=15 be equation no.1

then the line which is parallel to the equation 1st

2 x+5y+k = 0 let it be equation no.2

now the equation no.2 passes through the point (-10,1)

or, 2x+5y+k =0

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or, -20+5+k= 0

or, -15+k= 0

or, k= 15

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or, 2x+5y+k=0

or, 2x+5y+15=0

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1 year ago
Please give me the the answer of the question in the link that's 50 points
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Step-by-step explanation:

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Find the reduced row echelon form of the following matrices and then give the solution to the system that is represented by the
GaryK [48]

Answer:

a)

Reduced Row Echelon:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]

Solution to the system:

x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}

b)

Reduced Row Echelon:

\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]

Solution to the system:  

x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

Step-by-step explanation:

To find the reduced row echelon form of the matrices, let's use the Gaussian-Jordan elimination process, which consists of taking the matrix and performing a series of row operations. For notation, R_i will be the transformed column, and r_i the unchanged one.

a) \left[\begin{array}{cccc}0&4&7&0\\2&1&0&0\\0&3&1&-4\end{array}\right]

Step by step operations:

1. Reorder the rows, interchange Row 1 with Row 2, then apply the next operations on the new rows:

R_1=\frac{1}{2}r_1\\R_2=\frac{1}{4}r_2

Resulting matrix:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&3&1&-4\end{array}\right]

2. Set the first row to 1

R_3=-3r_2+r_3

Resulting matrix:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]

3. Write the system of equations:

x_1+\frac{1}{2}x_2=0\\x_2+\frac{7}{4}x_3=0\\x_3=-4

Now you have the  reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}

b)

\left[\begin{array}{cccc}4&3&0&7\\8&6&2&-3\\4&3&2&-10\end{array}\right]

1. R_2=-2r_1+r_2\\R_3=-r_1+r_3

Resulting matrix:

\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]

2. Write the system of equations:

4x_1+3x_2=7\\2x_3=-17

Now you have the reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

7 0
3 years ago
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