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2 years ago

When a number is decreased by 40% of itself the result is 96. What is the number?

Mathematics

2 answers:

Natalija [7]2 years ago

5 0

Q) 96/(100% - 40%)

→ 96/ 60%

→ 96/ 0.6

→ 160 is the number.

Nikitich [7]2 years ago

3 0

Answer:

160

Step-by-step explanation:

96 / (100%-40%) = 96/ (60%)

= 96/0.6 = 160

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3 years ago

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Find dy/dx any one <br>x=acost , y=asint

dangina [55]

Answer:

$x=a \cos t \implies \dfrac{dx}{dt}=-a \sin t$

$y= a \sin t \implies \dfrac{dy}{dt}=a \cos t$

Using the chain rule:

$\begin{aligned}\implies \dfrac{dy}{dx} & =\dfrac{dy}{dt} \times \dfrac{dt}{dx}\\\\& = a \cos t \times \dfrac{1}{-a \sin t}\\\\& = \dfrac{a \cos t}{-a \sin t}\\\\& = - \dfrac{\cos t}{\sin t}\\\\& = - \cot t\end{aligned}$

8 0

2 years ago

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:

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

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

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

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

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

Finally,

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

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

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

3 0

3 years ago