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tia_tia [17]
2 years ago
10

From a plane flying due east at 265m above above sea level, the angle of depression from the plane to a ship is 25 degrees. How

far is the plane from the ship? Round to the nearest tenth of a foot.
Mathematics
1 answer:
Artyom0805 [142]2 years ago
8 0
627.0m the process is shown in the following picture.

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Yakvenalex [24]

Answer: 10 < x < 60, or x is from 10 to 60

Step-by-step explanation: A domain of a graph consists of all of the input values shown on the x-axis while the range consists of all of the output values shown on the y-axis. Since the question asks to find the domain, you should look at the values of the graph from left to right. The domain includes all real numbers. Notice the black horizontal line on the graph and match up the beginning of it and the end of it with the values on the x-axis. The line begins at an x-value of 10 and ends at an x-value of 60; therefore, the domain (x-values) is from 10 to 60.

8 0
3 years ago
Solve for x<br> type your answer as x = (#)<br> ASAP
Nataliya [291]

Answer:

One step Equations:

1. x=4

2. x=18

3. x=3

4. ×=50

Two Step Equations:

1. x= 7

2. x= -4

3. x= 24

Variations on Both Sides:

(I'm not sure if these are right but the answers above are for sure correct)

1. x= 4.5

2. x= -2

3 0
3 years ago
Read 2 more answers
A club offers me membership for a fee of 20$ plus a monthly fee of 18 b club offers a fee of 14$ and a monthly fee of 20 after h
MArishka [77]

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Step-by-step explanation:

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4 0
3 years ago
Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3
nlexa [21]

Answer:

(x,y) = (5,-6)

Step-by-step explanation:

\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\

                     ~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)

\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\

\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5

y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6

\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)

7 0
2 years ago
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