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Nikitich [7]
3 years ago
9

Toucans and​ blue-and-yellow macaws are both tropical birds. The length of the toucan in the picture to the right is about two t

hirds of the length of the​ blue-and-yellow macaw. The toucan is about 18 in long. What is the length of the​ blue-and-yellow macaw?
The length of the​ blue-and-yellow macaw is about __in.
Mathematics
2 answers:
docker41 [41]3 years ago
7 0

Answer:

The length of blue-and-yellow macaw is 27 inches.

Step-by-step explanation:

The length of the toucan is about two thirds of the length of the​ blue-and-yellow macaw.

The toucan is about 18 in long.

Let the length of blue-and-yellow macaw be x.

We know that 2/3 of x is equal to 18.

We get this as:

\frac{2x}{3} =18

=> 2x=54

x = 27

Hence, the length of blue-and-yellow macaw is 27 inches.

Eddi Din [679]3 years ago
5 0

the length in in. will be 27


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Find the missing length indicated. Leave your answer in simplest radical form.
pickupchik [31]
25

The triangle given by 9, unknown, and 15 and the triangle given by 15, unknown, and x are similar triangles and therefore 9:15 = 15:x

X is 25 in this case
8 0
2 years ago
the fire hydrant sits 72 feet from the base of a 125-foot buiding. Find the ANGLE OF ELEVATION from the fire hydrant to the top
PSYCHO15rus [73]

Answer:

  60°

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relationship between angles and sides in a right triangle. Here, we're given the side Adjacent and the side Opposite the angle of interest. This makes it appropriate to choose the relation ...

  Tan = Opposite/Adjacent

For our angle of elevation α, this means ...

  tan(α) = 125/72

  α = arctan(125/72) = 60.058°

  α ≈ 60°

The angle of elevation is about 60°.

5 0
3 years ago
Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

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<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
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8 0
3 years ago
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deff fn [24]

Answer:

last option

Step-by-step explanation:

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5 0
3 years ago
Divide 4^2/4^6
deff fn [24]

Step-by-step explanation:

=  {4}^{2}  \div  {4}^{6}

=  {4}^{2 - 6}

=  {4}^{ - 4}

Option → C

\:

=  {4}^{ - 4}

=  \frac{1}{ {4}^{4} }

=  \frac{1}{4 \times 4 \times 4 \times 4}

=  \frac{1}{256}

Option → A

6 0
3 years ago
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