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Alenkinab [10]
3 years ago
9

Please tell me the angle of elevation

Mathematics
1 answer:
nikitadnepr [17]3 years ago
8 0

Answer:

32°

Step-by-step explanation:

The model forms a triangle. The sum of the angles of a triangle is 180°. Solve with the following equation:

? + 90 + 58 = 180

? + 148 = 180

? = 32

Therefore, the angle of elevation is 32°.

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Karen walked faster than Bob

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2 years ago
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My name is Ann [436]

Answer:

25, -13, -6.3, -7.8, -1.9, 2, 7.8

Step-by-step explanation:

5 0
3 years ago
which if the following describes the transformations of g(x)=-(2)^x+4 -2 from the parent function f(x)=2^x?
Pie
Compairing g(x) with f(x), we can write:

To get g(x) f(x) is:
1) Multiplied by -1.
This implies the f(x) is reflected over x-axis

2) Addition of 4 to x
This implies f(x) is shifted 4 units to left

3) Subtraction of 2 to function value.
This implies f(x) is shifted down by 2 units.

So, we can write g(x) as:

g(x) = -f(x+4) - 2

So, the correct answer to this question is option A.
Shift 4 units left, reflect over x axis and shift 2 units down
6 0
3 years ago
Read 2 more answers
Quadrilateral abcd is inscribed in a circle with m∠a = (x)degrees, m∠b = (2x - 10)degrees, and m∠c = (3x)degree. what is m∠d?
Lynna [10]
Hello,

When you have a polygon inscribed in a circle, the opposite angles are supplementary.

So we can start by saying that A + C = 180. Or x + 3x = 180. This gives us a value of 45 for x.

If we input that in for angle B, we get: 2(25) - 10 = 80.

If angle B = 80, then angle D must be 100, because the angles are supplementary.

Good luck,
MrEQ
6 0
4 years ago
If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1<br><br><br> help meeeee plessss ​
Trava [24]

A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}

where det(<em>A</em>) = 1×1 - 2×1 = -1.

B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}

where det(<em>B</em>) = 0×2 - (-1)×1 = 1. Then

B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

On the other side, we have

AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}

and det(<em>AB</em>) = det(<em>A</em>) det(<em>B</em>) = (-1)×1 = -1. So

(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

and both matrices are clearly the same.

More generally, we have by definition of inverse,

(AB)(AB)^{-1} = I

where I is the identity matrix. Multiply on the left by <em>A </em>⁻¹ to get

A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}

Multiplication of matrices is associative, so we can regroup terms as

(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}

Now multiply again on the left by <em>B</em> ⁻¹ and do the same thing:

B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}

7 0
3 years ago
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