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4 years ago

Solve for x. Please show work.

Mathematics

1 answer:

irakobra [83]4 years ago

6 0

Answer:

First exercise: $x=7$

Second exercise: $x=2$

Step-by-step explanation:

Acording to the Intersecting Secants Theorem the products of the segments of two secants that intersect each other outside a circle, are equal.

Based on this, in order to solve the first exercise and the second exercise, we can write the following expressions and solve for "x":

First exercise:

$(5)(5+x)=6(6+4)\\\\25+5x=60\\\\5x=60-25\\\\x=\frac{35}{5}\\\\x=7$

Second exercise:

$(4)(4+x)=3(3+5)\\\\16+4x=24\\\\4x=24-16\\\\x=\frac{8}{4}\\\\x=2$

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HIJK is a rectangle. Find the value of x and find the length of each diagonals. If hj=19+2x and ik=3x+22

viva [34]

Hj and ik are opposite sides of the rectangle so they areequal in length.

19 + 2x = 3x + 22
19-22 = 3x-2x
x = -3

We can't find the length of the diagonals as we have no information about the width.

8 0

4 years ago

The figures are similar. Give the ratio of the perimeters and the ratio of the areas of the first figure to the second.

pickupchik [31]

Answer: first option.

Step-by-step explanation:

The ratio of the area of the triangles can be calculated as following:

$ratio_{(area)}=(\frac{l_1}{l_2})^2$

Where $l_1$ is the lenght of the given side of the smaller triangle and $l_2$ is the lenght of the given side of the larger triangle.

Therefore:

$ratio_{(area)}=(\frac{28}{32})^2=\frac{49}{64}$

It can be written as following:

$ratio_{(area)}=49:64$

The ratio of the perimeter is:

$ratio_{(perimeter)}=\frac{l_1}{l_2}$

Where $l_1$ is the lenght of the given side of the smaller triangle and $l_2$ is the lenght of the given side of the larger triangle.

Therefore:

$ratio_{(perimeter)}=\frac{28}{32}=\frac{7}{8}$

It can be written as following:

$ratio_{(perimeter)}=7:8$

5 0

4 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$