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lyudmila [28]
3 years ago
12

Find w and y without a calculator, will give brainliest for the correct answer

Mathematics
1 answer:
love history [14]3 years ago
6 0

Answer: w=4, y=4

Step-by-step explanation:

For this problem, we can use the 30-60-90 triangle to find out what the length of w and y are. 30-60-90 triangle is a special triangle. The hypotenuse is 2x in length. It is directly opposite the right angle. The leg opposite of 60° is x√3 in length. The leg opposite of 30° is x. For all 3 legs, wherever you see x, you plug in the same number.

We can look at the figure as 2 separate triangles.

For the triangle on the right, we can see the hypotenuse is 8. Since we know the length of the hypotenuse is 2x, we can plug in 8 for 2x to find x.

2x=8

x=4

Now that we know x=4, we can directly plug it into the lengths above.

W is across from 30°. Above, we have established that the leg across from 30° has the length of x. Since x=4, w=4.

Since w=4, we can use this information to find the length of y by looking on the left triangle. Now, y is across from 30°. In the first paragraph, we stated that the leg across from 30° is x. Since we know x, we can directly plug it into this. After we plug it in, y=4.

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  77 purple counters

Step-by-step explanation:

There are a total of 5+7=12 ratio units, so each stands for 132/12 = 11 counters. The 7 ratio units representing the purple counters stand for ...

  11 × 7 = 77 purple counters

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Solve:<br> |x -0.5| &lt; 12
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Step-by-step explanation:

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Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

7 0
4 years ago
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