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AURORKA [14]
2 years ago
14

Find the greatest common monomial factor : 4a^5b^2 and ab

Mathematics
1 answer:
3241004551 [841]2 years ago
8 0
4a^5b^2=2*2*a*a*a*a*a*b*b
ab=a*b
greatest factor is ab
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SOS!!
wlad13 [49]

The only difference between the two following expressions is their exponent though it’s equal. If you would analyze carefully, the two expressions are equal since 2/4 is just equal to ½. The two expressions are equal in value so the presentation of the exponent is their only difference.

4 0
3 years ago
(8-9)-(9+3)*5<br> 5+(5+5+6)+9<br> (5/3-3)<br> 2*3+(4*2)/2
pashok25 [27]

Answer:

If you need any further help, feel free to let me know

Step-by-step explanation:

The answers are

(8-9)-(9+3)*5=-61

5+(5+5+6)+9=30

(5/3-3)= -4 over 3 or -4/3 or -1.3 or -1 1/3 depends on what your choices are but all those are correct

2*3+(4*2)/2=10

6 0
2 years ago
Use the value of the linear correlation coefficient r to find the coefficient of determination and the percentage ofthe total va
Vikentia [17]

Answer:

Step-by-step explanation:

The coefficient of determination = r^{2}  = 0.885^{2} = 0.7832

It means about 78% variation in waist size of males can be explained by their weight and about 23% can not be explained.

4 0
3 years ago
Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein
ivanzaharov [21]

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min

Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

dV/dt = 1515327.582 cm³/min

Since dV/dt = k - 6800 cm³/min

k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min

5 0
3 years ago
Somebody please help me with this
kap26 [50]

Answer:

D. $7/hour

Step-by-step explanation:

128-86=42

42/6=7

6 0
2 years ago
Read 2 more answers
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