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3 years ago

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Lin is making two pans of brownies. She lines the bottom of her first pan with aluminum foil. The area of the rectangular piece

of foil is 11 and 1/4 sq inches. Its length is 4 and 1/2 inches. What is the width of the piece of foil?

1 answer:

Gemiola [76]3 years ago

6 0

Answer: 2 1/2 inches

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What is 8 divided by 0.24

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The answer is 33.3333

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You make candles by adding 2 fluid ounces of scented oil for every 22 fluid ounces of wax. Your friend makes candles by adding 3

Rina8888 [55]

Answer:

Your candles are more fragrant. ;)

Step-by-step explanation:

2/22 > 3/37

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2 years ago

One way of solving systems of linear equations is by adding a multiple of one equation to the other. the multiplier for one equa

elena-s [515]

We want to use elimination to solve
y = 0 (1)
x + y = 40 (2)

Multiply (1) by -1 to eliminate y.
-y = 0 (3)

Add (2) and (3).
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3 years ago

What is the answer to <br> 6+0.10x=0.15x+8

asambeis [7]

Let's solve this problem step-by-step.

6+0.1x=0.15x+8

Step 1: Simplify both sides of the equation.

0.1x+6=0.15x+8

Step 2: Subtract 0.15x from both sides.

0.1x+6−0.15x=0.15x+8−0.15x

−0.05x+6=8

Step 3: Subtract 6 from both sides.

−0.05x+6−6=8−6

−0.05x=2

Step 4: Divide both sides by -0.05.

-0.05x/-0.05=2/-0.05

So, the answer for this problem is x=-40

4 0

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

<u>2x-2y=-6 </u>

<u>7x =-21</u>

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)]

3 0

3 years ago