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Nikitich [7]
3 years ago
9

Find the perimeter of ABCD please.

Mathematics
1 answer:
Komok [63]3 years ago
5 0

Answer:

Perimeter = 71

Step-by-step explanation:

angles formed by tangent lines has equal side lengths.

segments from A to tangent point = 17 each

segments from B to tangent point = 12 each  (29 - 17 = 12)

segments from C to tangent point = 2.5 each

segments from D to tangent point = 4 each  (21 - 17 = 4)

Perimeter = 2(17) + 2(12) + 2(2.5) + 2(4) = 71

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One side of a rectangle is three times as long as another. If its area is 363 square centimeters, find its dimensions.
Yakvenalex [24]

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  11 cm by 33 cm

Step-by-step explanation:

You can solve this problem mentally as follows.

Consider the rectangle as 3 squares, side-by-side. Then the area of each of those squares is 363/3 = 121 cm^2. From your knowledge of the squares of numbers, you know that 121 = 11^2. So, the width of the rectangle is 11 cm, and the length is 3 times that, or 33 cm.

_____

Using variables, we can let w represent the width. Then 3w can represent the length, and the area is ...

  A = LW

  A = (3w)(w) = 3x^2 = 363

  w^2 = 363/3 = 121

  w = √121 = 11

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The width is 11 cm; the length is 33 cm.

7 0
3 years ago
Read 2 more answers
-6/5-2/5v+4/15+1/3v please help
PtichkaEL [24]

Answer:

-14/15-1/15v

Step-by-step explanation:

First you make all of the fractions have the common denominator 15.

And then you have -18/15-6/15v+4/15+5/15v

Then you combine like terms to get -14/15-1/15v

5 0
3 years ago
(x+7)/(x^2-49) find the domain. show work
harina [27]
\large\begin{array}{l} \textsf{Find the domain of}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-49}}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-7^2}}\\\\\\ \textsf{Factor out the denominator using special products:}\\\\ \textsf{(a difference of squares)}\\\\ \mathsf{f(x)=\dfrac{x+7}{(x+7)(x-7)}} \end{array}


\large\begin{array}{l} \textsf{Restrictions for the domain:}\\\\ \bullet~~\textsf{Denominators must not be zero:}\\\\ \mathsf{(x+7)(x-7)\ne 0}\\\\ \begin{array}{rcl} \mathsf{x+7\ne 0}&~\textsf{ and }~&\mathsf{x-7\ne 0}\\\\ \mathsf{x\ne -7}&~\textsf{ and }~&\mathsf{x\ne 7} \end{array} \end{array}


\large\begin{array}{l} \textsf{Therefore, the domain of f is}\\\\ \mathsf{D_f=\{x\in\mathbb{R}:~~x\ne -7~~and~~x\ne 7\}}\\\\\\ \textsf{or using a more compact form}\\\\ \mathsf{D_f=\mathbb{R}\setminus\{-7,\,7\}}\\\\\\ \textsf{or using the interval notation}\\\\ \mathsf{D_f=\left]-\infty,\,-7\right[\,\cup\,\left]7,\,+\infty\right[.} \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2155752


\large\textsf{I hope it helps. :-)}



Tags: <em>function domain real rational factorizing special product interval</em>

</span>
7 0
3 years ago
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