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olya-2409 [2.1K]

3 years ago

15

20 points and brainliest!!!! Pls help er

1 answer:

tensa zangetsu [6.8K]3 years ago

3 0

Slader could really help

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Use graphs and tables to find the limit and identify any vertical asymptotes of the function. (8 points)

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4 years ago

Read 2 more answers

A 2.00 kg particle has the xy coordinates (−1.20 m, 0.500 m), and a 4.00 kg particle has the xy coordinates (0.600 m, −0.750 m).

Dafna11 [192]

Answer:

$(x_{3},y_{3})=(-1.50\ m, -1.43\ m)$

Step-by-step explanation:

Given:

Mass of the first particle $m_{1} =2.0\ kg$

Mass of the second particle $m_{2} =4.0\ kg$

Mass of the third particle $m_{3} =\ ?$

xy coordinate of the first particle $(x_{1},y_{1})=(-1.20 m, 0.500 m)$

xy coordinate of the second particle $(x_{2},y_{2})=(0.600 m, -0.750 m)$

coordinate of the centre of mass of three-particle system $(x_{cm},y_{cm})=(-0.500 m, -0.700 m)$

We need to find the xy coordinate of the third particle.

Solution:

First we calculate total mass of the three-particle system.

$M = m_{1}+m_{2}+m_{3}$

$M = 2.0+4.0+3.0$

$M = 9.0\ kg$

x coordinate of the centre of mass of the three-particle system are given as;

$x_{cm}=\frac{1}{M}(m_{1}x_{1}+ m_{2}x_{2}+ m_{3}x_{3})$

Rewrite the equation for $x_{3}$

$m_{3}x_{3}=Mx_{cm}-m_{1}x_{1}-m_{2}x_{2}$

Part a.

$x_{3}=\frac{1}{m_{3}} (Mx_{cm}-m_{1}x_{1}-m_{2}x_{2})$

Substitute all given values in above equation.

$x_{3}=\frac{1}{3.0} (9.0\times (-0.5)-2.0\times (-1.20)-4.0\times (0.6))$

$x_{3}=-1.50\ m$

Part b.

Similarly for $y_{3}$ coordinate.

$y_{3}=\frac{1}{m_{3}} (My_{cm}-m_{1}y_{1}-m_{2}y_{2})$

Substitute all given values in above equation.

$y_{3}=\frac{1}{3.0} (9.0\times (-0.7)-2.0\times 0.5-4.0\times (-0.75))$

$y_{3}=-1.43\ m$

Therefore, xy coordinate of the third particle $(x_{3},y_{3})=(-1.50\ m, -1.43\ m)$

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3 years ago

5. In an online practice test, Jack scores 18 points for the

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Answer 12
Step-by-step explaining
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An example of a cost would be _____.

inna [77]

Answer:

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Step-by-step explanation:

Its cost means something about money

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3 years ago

Find the derivative r '(t) of the vector function r(t). t sin 6t , t2, t cos 7t Part 1 of 4 The derivative of a vector function

stich3 [128]

Answer:

$r'(t)=(sin(6t)+6tcos(6t),2t,cos(7t)-7tsin(7t))$

Step-by-step explanation:

We need to find the derivative $r'(t)$ of the vector function :

$r(t)=(tsin(6t),t^{2},tcos(7t))$

In order to find $r'(t)$ , we are going to differentiate each of its components ⇒

We can write the following ⇒

$r(t)=(f(t),g(t),h(t))=(tsin(6t),t^{2},tcos(7t))$ ⇒

$f(t)=tsin(6t)\\g(t)=t^{2}\\h(t)=tcos(7t)$

Let's differentiate each function to obtain $r'(t)$ :

$f(t)=tsin(6t)$ ⇒ $f'(t)=1.sin(6t)+t.cos(6t).6=sin(6t)+6tcos(6t)$ ⇒

$f'(t)=sin(6t)+6tcos(6t)$

Now with $g(t)$ :

$g(t)=t^{2}$ ⇒

$g'(t)=2t$

With $h(t)$ :

$h(t)=tcos(7t)$ ⇒ $h'(t)=1.cos(7t)+t[-sin(7t)].7$ ⇒

$h'(t)=cos(7t)-7tsin(7t)$

Finally we need to complete $r'(t)=(f'(t),g'(t),h'(t))$ with its components :

$r'(t)=(sin(6t)+6tcos(6t),2t,cos(7t)-7tsin(7t))$

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3 years ago