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borishaifa [10]

3 years ago

14

The weight V of an object on Venus varies directly with its weight E on Earth. A person weighing 120 lb on Earth would weigh 106

lb on Venus. How much would a person weighing 150 lb on Earth weigh on Venus?

2 answers:

sukhopar [10]3 years ago

5 0

We have
120lbs Earth................................106lbs Venus
150lbs Earth...................................?Venus

Venus weight=150*106/120=132.50 lbs

svlad2 [7]3 years ago

4 0

Answer:

The person weighing 150 lb on Earth will weight 132.5 lb on Venus.

Step-by-step explanation:

The weight of object on Venus varies directly with its weight on the Earth.

$w_v\propto w_e$ ..(1)

$\frac{w_v}{w_e}=k=constant$

$w_v$ = Weight on Venus

$w_e$ = Weight on Earth

If the weight of person on Earth = $w_e=120 lb$

If the weight of that person on Venus= $w_v=106 lb$

And,the weight of another person on Earth = $w_e'=150 lb$

The weight of another person on Venus= $w_v'=?$

$\frac{w_v}{w_e}=\frac{w_v'}{w_e'}$ (From 1)

$w_v'=\frac{w_v\times w_e'}{w_e}=\frac{106 lb\times 150 lb}{120 lb}$

$w_v'=132.5 lb$

The person weighing 150 lb on Earth will weight 132.5 lb on Venus.

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3 years ago

The polar coordinates of a point are 3π 4 and 7.00 m. What are its Cartesian coordinates (in m)? (x, y) = 6.06,−3.5 m

Keith_Richards [23]

Answer:

a) $\left(x,y\right)=\left(4.95,-4.95\right)$

b) $r\angle\theta = 7\angle0.5236\,\text{radians}$

Step-by-step explanation:

Polar coordinates are represented as: $r\angle\theta$ , where 'r' is the length (or magnitude) of the line, and ' $\theta$ ' is the angle measured from the positive x-axis.

in our case:

$7\angle\dfrac{3\pi}{4}$

to covert the polar to cartesian:

$x = r\cos{\theta}$

$y = r\sin{\theta}$

we can plug in our values:

$x = 7\cos{\dfrac{3\pi}{4}} = -7\dfrac{\sqrt{2}}{2}$

$y = 7\sin{\dfrac{3\pi}{4}} = 7\dfrac{\sqrt{2}}{2}$

the Cartesian coordinates are:

$\left(x,y\right)=\left(-7\dfrac{\sqrt{2}}{2},7\dfrac{\sqrt{2}}{2}\right)$

$\left(x,y\right)=\left(4.95,-4.95\right)$

(b) to convert (x,y) = (6.06,-3.5)

we'll use the pythagoras theorem to find 'r'

$r^2 = x^2+y^2$

$r^2 = (6.06)^2+(-3.5)^2$

$r = \sqrt{48.97} \approx 7$

the angle can be found by:

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

$\tan{\theta} = \dfrac{3.5}{6.06}$

$\theta = \arctan{left(\dfrac{3.5}{6.06}\right)}$

$\theta = 0.5236 \text{radians}$

to convert radians to degrees:

$\theta = 0.5236 \times \dfrac{180}{\pi} \approx 30^\circ$

writing in polar coordinates:

$r\angle\theta = 7\angle30^\circ\,\,\text{OR}\,\,7\angle0.5236\,\text{radians}$

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Answer:

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2 years ago

In circle O, m∠R = 30.8°. Find m∠NOQ.

nordsb [41]

Given:

In circle O, m∠R = 30.8°.

To find:

The m∠NOQ

Solution:

Central angle theorem: According to this theorem, the central angle is always twice of subtended angle on the same arc.

Angle NOQ and angle NRQ are on the same arc but ∠NOQ is the central angle and ∠NRQ is the subtended angle on the arc NQ.

Using central angle theorem, we get

$m\angle NOQ=2\times m\angle NRQ$

$m\angle NOQ=2\times m\angle R$

$m\angle NOQ=2\times 30.8^\circ$

$m\angle NOQ=61.6^\circ$

Therefore, $m\angle NOQ=61.6^\circ$ .

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aleksklad [387]

HEY mate here is pratyush from India for your help...

the answer is with the attachment

hope it helps you..

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3 years ago

Read 2 more answers