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2 years ago

GeometryNeed help finding A and B

Mathematics

1 answer:

MariettaO [177]2 years ago

4 0

Answer:

a = 8

b = 4√5

Step-by-step explanation:

Corresponding sides of similar triangles are proportional.

short side/long side = a/16 = 4/a

a² = 64 . . . . . . multiply by 16a

a = 8 . . . . . . take the square root

hypotenuse/short side = b/4 = 20/b

b² = 80 . . . . . . . multiply by 4b

b = 4√5 . . . . . take the square root

_____

Additional comment

These relations for the right triangle can be summarized by the rule "the length of a segment that touches the hypotenuse is the geometric mean of the two hypotenuse segments it touches." In the case of b, that means ...

b = √((4)(4+16)) = 4√5

The same sort of relation applies to the unmarked vertical segment on the left:

c = √((16)(16+4)) = 8√5

And, we saw ...

a = √((4)(16)) = 8

Of course, if this rule is to make sense, you need to know that the geometric mean of two numbers is the square root of their product. (In general, the geometric mean of n numbers is the n-th root of their product.)

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3 years ago

A = ???? 4 −2

liraira [26]

Answer:

1. $A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{1}{5}\\\frac{1}{10}&\frac{2}{5} \end{array}\right]$

2. $|B|=0$

Step-by-step explanation:

We have the matrix:

$A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]$

It's a 2×2 matrix (This means that the matrix has two rows and two columns).

1. We have to find the inverse of A.

For a 2×2 matrix the inverse is:

If you have $A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

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

And,

$|A|$ is the determinant of the matrix, the determinant has to be different from zero.

If $|A|=0$ then the matrix doesn't have inverse.

$|A|=ad-bc$

Then,

$A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]$

$a=4, b=-2, c=-1, d=3$

First we are going to calculate the determinant:

$|A|=ad-bc\\|A|=4.3-(-2).(-1)=12-2\\|A|=10$

The determinant is different from zero, then the matrix has inverse.

Then the inverse of A is:

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

$A^-^1=\frac{1}{10} \left[\begin{array}{cc}3&-(-2)\\-(-1)&4\end{array}\right]\\\\\\A^-^1=\frac{1}{10} \left[\begin{array}{cc}3&2\\1&4\end{array}\right]\\\\\\A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{2}{10}\\\frac{1}{10}&\frac{4}{10} \end{array}\right]\\\\\\A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{1}{5}\\\frac{1}{10}&\frac{2}{5} \end{array}\right]$