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Lilit [14]
3 years ago
12

Masha is making a scale diagram of the front face of an aquarium. The face is shaped like a rectangle.

Mathematics
2 answers:
vlada-n [284]3 years ago
5 0

Answer:   28 centi

Step-by-step explanation:

Marat540 [252]3 years ago
4 0
Scale:
1 cm : 6 inches

---------------------------------
Length of the aquarium:
---------------------------------
5 feet = 5 x 12 = 60 inches
60 ÷ 6 = 10 
The length will be 10 cm on the scale drawing.

---------------------------------
Width of the aquarium:
---------------------------------
2 feet = 2 x 12 = 24 inches
24 ÷ 6 = 4 

The width will be 4 cm on the scale drawing.

---------------------------------
Find Perimeter:
---------------------------------
Perimeter = 2 (length + width)
Perimeter = 2( 10 + 4 )
Perimeter = 28 cm

--------------------------------------------------------------------------------------------------
Answer:  The perimeter of the scale drawing of the front face is 28 cm.--------------------------------------------------------------------------------------------------

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Polar and cartesian equation<h2>Initial explanation</h2>

Let's analyze the relation between r and x and y:

We have that between the indicated value of r (of the polar coordinates) and x and y (of the cartesian coordinates) there is a relation because they form a triangle. If r changes, then the value of x and y will change.

<h2>STEP 1: given equation</h2>

Using the given equation

r = 2 secØ

we have that

\begin{gathered} r=2secØ \\ \downarrow \\ \frac{r}{2}=secØ \end{gathered}<h2>STEP 2: secØ equation</h2>

Observing the image of the initial explanation we have a right triangle, we know that the equation of

secØ for any right triangle is given by:

\sec Ø=\frac{\text{hypotenuse}}{\text{adjacent side}}

In this case,

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adjacent side = x

then,

\begin{gathered} \sec Ø=\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{r}{x} \\ \sec Ø=\frac{r}{x} \end{gathered}<h2>STEP 3: comparison between given equation and secØ equation</h2>

Then, we have that:

\begin{gathered} \sec Ø=\frac{r}{x} \\ \sec Ø=\frac{r}{2} \end{gathered}

This means that:

\begin{gathered} \frac{r}{x}=\sec Ø=\frac{r}{2} \\ \downarrow \\ \frac{r}{x}=\frac{r}{2} \end{gathered}

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kobusy [5.1K]

Answer:

\huge\boxed{3 = 57 \textdegree}

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Step-by-step explanation:

When two angles are complementary, their angle lengths add up to 90°.

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Now we can solve for x.

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