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3 years ago

12

When Tara, a nurse, works on Saturdays, she is paid a $30 bonus plus $20 per hour worked.

1 answer:

erik [133]3 years ago

6 0

Too much work…. Sorry

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e-lub [12.9K]

Hope this helps
G=4.

3 0

2 years ago

What is 1 and 1/6 minus 7/12 ?

Damm [24]

$1 \frac{1}{6} - \frac{7}{12} \\ 
 \\ 1 \frac{1}{6}= \frac{1x6+1}{6}= \frac{7}{6} \\ 
 \\ \frac{7x2}{6x2}= \frac{14}{12} \\ 
 \\ \frac{7x1}{12x1}= \frac{7}{12} \\ 
 \\ \frac{14}{12} - \frac{7}{12}= \frac{7}{12} \\ 
 \\ Solution: \frac{7}{12} 
$

7 0

3 years ago

ILL BRAINLIEST YOU PLEASE HELP ME

liubo4ka [24]

15.8 I believe. Since 20 isn’t a option 15 is the closest. Good luck I have good faith I have you the right answer

8 0

3 years ago

Greg needs $51 to go on a field trip. He has saved $12.75. He earns $6.50 per hour cleaning his neighbor's garden, and he earns

Schach [20]

B.)Yes, because the total will be $52

5 0

3 years ago

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre

Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

$N=2.85 rev/s$

angular velocity $\omega =2\pi N=17.90 rad/s$

mass of objects $m=1.5 kg$

distance of objects from stool $r_1=0.789 m$

Combined moment of inertia of stool and student $=5.53 kg.m^2$

Now student pull off his hands so as to increase its speed to 3.60 rev/s

$\omega _2=2\pi N_2$

$\omega _2=2\pi 3.6=22.62 rad/s$

Initial moment of inertia of two masses $I_0=2mr_^2$

$I_0=2\times 1.5\times (0.789)^2=1.867$

After Pulling off hands so that r' is the distance of masses from stool

$I_0'=2\times 1.5\times (r')^2$

Conserving angular momentum

$I_1\omega =I_2\omega _2$

$(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62$

$I_0'=1.397\times 0.791$

$I_0'=5.851$

$5.53+2\times 1.5\times (r')^2=5.851$

$2\times 1.5\times (r')^2=0.321$

$r'^2=0.107009$

$r'=0.327 m$

7 0

3 years ago