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svetlana [45]
3 years ago
6

IMPORTANT. LOTS OF POINTS. SOLVE NOW. Rays OM , ON , and OP coincide at time t=0. At the same instant in time, OM and OP begin t

o rotate in the plane about point O, but in opposite directions. Ray ON remains fixed. Ray OM rotates at a constant rate of 3° per second with respect to ON , while OP rotates at a constant rate of 6° per second. What is the smallest positive value of t, in seconds, in which rays OM and OP will coincide again? When t=30 min, how many more revolutions has ray OP completed than ray OM ?
Mathematics
2 answers:
AURORKA [14]3 years ago
8 0

Answer:

t=40 seg OP does 15 more revolutions than OM

Step-by-step explanation:

i didn't do this, just answering it just in case someone doesnt look at the comments

Anton [14]3 years ago
6 0
I believe if Socratic is right the answer is -x+4y=8
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-3x - y = 15<br> y = -8x<br> (x, y) = (0<br> )<br> 9,0
Nataly [62]
<h3>The solution is (x, y) = (3, -24)</h3>

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

-3x - y = 15 -------- eqn 1

y = -8x ------ eqn 2

We have to find solution of (x, y)

We can solve by substitution method

<em><u>Substitute eqn 2 in eqn 1</u></em>

-3x - (-8x) = 15

-3x + 8x = 15

5x = 15

Divide both sides by 5

<h3>x = 3</h3>

Substitute x = 3 in eqn 2

y = -8(3)

<h3>y = -24</h3>

Thus solution is (x, y) = (3, -24)

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