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aev [14]
3 years ago
14

PLEASE HELP I NEEED ANSWER.. ASAP

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

Answer:

Step-by-step explanation:

Triangle IGH is a right angle triangle.

From the given right angle triangle

GI represents the hypotenuse of the right angle triangle.

With m∠I as the reference angle,

HI represents the adjacent side of the right angle triangle.

GH represents the opposite side of the right angle triangle.

To determine Hl, we would apply the tangent trigonometric ratio

Tan θ = opposite side/adjacent side. Therefore,

Tan 42 = 11/HI

0.9HI = 11

HI = 11/0.9

HI = 12.22

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are any two regular polygons similar. The area of the smaller figure is about 390 cm^2. choose the best estimate of the area of
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Answer:

The best estimate of the area of the larger figure is 693\ cm^{2}

Step-by-step explanation:

step 1

<em>Find the scale factor</em>

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If two figures are similar, then the ratio of its corresponding sides is equal to the scale factor

Let

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y-----> the corresponding side of the smaller figure

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z=\frac{x}{y}

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substitute

z=\frac{12}{9}=\frac{4}{3} -----> the scale factor

step 2

<em>Find the area of the larger figure</em>

we know that

If two figures are similar, then the ratio of its  areas is equal to the scale factor squared

Let

z----> the scale factor

x-----> the area of the larger figure

y-----> the area of the smaller figure

so

z^{2}=\frac{x}{y}

we have

z=\frac{4}{3}

y=390\ cm^{2}

substitute and solve for x

(\frac{4}{3})^{2}=\frac{x}{390}

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M1 Matrix is:
harkovskaia [24]

Answer:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

Step-by-step explanation:

Given matrix M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)

The Inverse of a 2X2 matrix

A=\left(\begin{array}{ccc}a&b\\c&d\end{array}\right)

can be found using the following:

A^{-1}=\dfrac{1}{ad-bc} \left(\begin{array}{ccc}d&-b\\-c&a\end{array}\right)

Therefore:

M^{-1}=\dfrac{1}{(5*-5)-(3*-8)} \left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=-1\left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)

Next, we find the product M^{-1}M

M^{-1}M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\\=\left(\begin{array}{ccc}-5*-5+3*-8&-5*3+3*5\\-8*-5+5*-8&-8*3+5*5\end{array}\right)\\=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)

Therefore:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

NOTE: The multiplication of a matrix and its inverse always gives the identity matrix as seen above,

7 0
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