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PilotLPTM [1.2K]

2 years ago

5

Wyatt’s eye-level height is 120 ft above sea level, and Shawn’s eye-level height is 270 ft above sea level. How much farther can

Shawn see to the horizon? Use the formula d = StartRoot StartFraction 3 h Over 2 EndFraction EndRoot, h greater-than-or-equal-to 0 with d being the distance they can see in miles and h being their eye-level height in feet.
StartRoot 5 EndRoot mi

3 StartRoot 5 EndRoot mi

15 StartRoot 5 EndRoot mi

45 StartRoot 5 EndRoot mi

1 answer:

earnstyle [38]2 years ago

3 0

Answer:

The answer is 3√5 mi.

The formula is: d = √(3h/2)

Wyatt:

h = 120 ft

d = √(3 * 120/2) = √180 = √(36 * 5) = √36 * √5 = 6√5 mi

Shawn:

h = 270 ft

d = √(3 * 270/2) = √405 = √(81 * 5) = √81 * √5 = 9√5 mi

How much farther can Shawn see to the horizon?

Shawn - Wyatt = 9√5 - 6√5 = 3√5 mi

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the scale on a map is stated as 1:250000 two villages are 9 cm apart on the map how far from each other are the two villages in

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2 years ago

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2 years ago

Read 2 more answers

Find c1 and c2 such that M2+c1M+c2I2=0, where I2 is the identity 2×2 matrix and 0 is the zero matrix of appropriate dimension.

Katyanochek1 [597]

The question is missing parts. Here is the complete question.

Let M = $\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$ . Find $c_{1}$ and $c_{2}$ such that $M^{2}+c_{1}M+c_{2}I_{2}=0$ , where $I_{2}$ is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.

Answer: $c_{1} = \frac{-16}{10}$

$c_{2}=\frac{-214}{10}$

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:

$\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

$M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$

$M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]$

Solving equation:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.

So, the equation is:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

And the system of equations is:

$6c_{1}+c_{2} = -31\\-4c_{1}+c_{2} = -15$

There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:

$6c_{1}+c_{2} = -31\\(-1)*-4c_{1}+c_{2} = -15*(-1)$

$6c_{1}+c_{2} = -31\\4c_{1}-c_{2} = 15$

$10c_{1} = -16$

$c_{1} = \frac{-16}{10}$

With $c_{1}$ , substitute in one of the equations and find $c_{2}$ :

$6c_{1}+c_{2}=-31$

$c_{2}=-31-6(\frac{-16}{10} )$

$c_{2}=-31+(\frac{96}{10} )$

$c_{2}=\frac{-310+96}{10}$

$c_{2}=\frac{-214}{10}$

<u>For the equation, </u> $c_{1} = \frac{-16}{10}$ <u> and </u> $c_{2}=\frac{-214}{10}$ <u />

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2 years ago