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3 years ago

15

G ( h + 3/5 ) = 11 solve for h

1 answer:

Alenkasestr [34]3 years ago

6 0

Answer:

h = 11/g - 3/5

Step-by-step explanation:

$g ( h + \frac{3}{5} ) = 11 \\\\Divide\:both\:sides\:of\:the\:equation\:by\:g \:;\\\frac{g(h+\frac{3}{5}) }{g} = \frac{11}{g} \\\\h + \frac{3}{5} = \frac{11}{g} \\\\Move\: 3/5\: to\: the\: right\:and\:change\:its\:sign\\h = \frac{11}{g} -\frac{3}{5}$

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A camera is mounted at a point 4000 feet away from a geyser. If the water is rising vertically at 900 ft/s when it is 3000 feet

Leno4ka [110]

Answer:

$\dot \theta = 0.144\,\frac{rad}{s}$ , $\dot \theta = 8.251\,\frac{deg}{s}$ (Option B)

Step-by-step explanation:

The trigonometric diagram is included herein as attachment. The expression is presented below:

$\tan \theta = \frac{y}{x}$

Where:

$x$ - Horizontal distance between the geyser and the camera.

$y$ - Vertical distance between the geyser and the camera.

The rate of change in terms of time is:

$\dot \theta \cdot \sec^{2}\theta = \frac{\dot y\cdot x-y\cdot \dot x}{x^{2}}$

$\dot \theta \cdot \left(\frac{1}{\cos^{2}\theta} \right) = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}}$

$\dot \theta = \left(\frac{\dot y \cdot x - y \cdot \dot x}{x^{2}} \right)\cdot \cos^{2}\theta$

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$\dot \theta = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}+y^{2}}$

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$\dot \theta = \frac{\left(900\,\frac{ft}{s} \right)\cdot (4000\,ft)-(3000\,ft)\cdot \left(0\,\frac{ft}{s} \right)}{(4000\,ft)^{2}+(3000\,ft)^{2}}$

$\dot \theta = 0.144\,\frac{rad}{s}$

$\dot \theta = 8.251\,\frac{deg}{s}$

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