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3 years ago

13

A table has an area of 12 ft squared and a perimeter of 16 ft. What are the dimensions of the table?

2 answers:

AysviL [449]3 years ago

8 0

Answer:

A, 2 ft by 6 ft

Step-by-step explanation:

Likurg_2 [28]3 years ago

7 0

P=2x+2y, y=(p-2x)/2

a=xy, using y found above in this equation gives you:

a=(px-2x^2)/2 we are told a=12 and p=16 so

12=(16x-2x^2)/2

24=16x-2x^2

2x^2-16x+24=0

2x^2-12x-4x+24=0

2x(x-6)-4(x-6)=0

(2x-4)(x-6)=0

2(x-2)(x-6)=0

So the table is 2 ft by 6 ft.

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3 years ago

Read 2 more answers

If A is nonsingular, then (AT )-1 =(A-1) T . (a) Verify this theorem for 2 × 2 matrices

kipiarov [429]

Answer:

Verified

Step-by-step explanation:

Let the 2x2 matrix A be in the form of:

$\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

Where det(A) = ad - bc # 0 so A is nonsingular:

Then the transposed version of A is

$A^T = \left[\begin{array}{cc}a&c\\b&d\end{array}\right]$

Then the inverted version of transposed A is

$(A^T)^{-1} = \frac{1}{ad - cb} \left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]$

The inverted version of A is:

$A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-b\\-c&d\end{array}\right]$

The transposed version of inverted A is:

$(A^{-1})^T = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]$

We can see that

$(A^T)^{-1} = (A^{-1})^T$

So this theorem is true for 2 x 2 matrices

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3 years ago

Can you step by step solve 20x+12=4x-16

Murljashka [212]

Answer:

x = - $\frac{7}{4}$

Step-by-step explanation:

Given

20x + 12 = 4x - 16 ( subtract 4x from both sides )

16x + 12 = - 16 ( subtract 12 from both sides )

16x = - 28 ( divide both sides by 16 )

x = $\frac{-28}{16}$ = - $\frac{7}{4}$ [ = - 1.75 ]

4 0

3 years ago