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xeze [42]
3 years ago
14

Determine the equation of the circle graphed below

Mathematics
1 answer:
nikdorinn [45]3 years ago
5 0
(x-0)^2+(y-4)^2=6^2
Hope this helps
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U
True [87]

Answer:

D. RWS and SWT

Step-by-step explanation:

Adjacent angles have a common side and a common vertex but they don't overlaps each other

The two angles that has these requirements are :

RWS and SWT

8 0
3 years ago
Read 2 more answers
6/11+1/2 what is the answer
tester [92]

Answer:

1 1/22

Step-by-step explanation:


5 0
3 years ago
Umm pls help me ik how to do it but not sure on this specific one
Lapatulllka [165]

9514 1404 393

Answer:

  -8, +9

Step-by-step explanation:

You need factors of (6)(-12) = -72 that have a sum of +1.

From your knowledge of multiplication tables, you know that 72 = 8·9. Then -72 = (-8)(9), and those two factors have a sum of +1.

  6x^2 +x -12

  = 6x^2 +9x -8x -12

  = 3x(2x +3) -4(2x +3)

  = (3x -4)(2x +3)

8 0
3 years ago
I need help majorly on questions 13 14 15 16 and 17. Please help now In a super big rush
Sholpan [36]

13- 8

17-19

16- 16


All I did was add them up you could also multiply by two because you could

5 0
3 years ago
A 12-m3 oxygen tank is at 17°C and 850 kPa absolute. The valve is opened, and some oxygen is released until the pressure in the
sweet-ann [11.9K]

Answer:

Released oxygen mass: 15.92 kg

Step-by-step explanation:

ideal gas law : P*V=nRT

P:pressure

V:volume

T:temperature

n:number of moles of gas

n [mol] = m [g] /M [u]

m : masa

M: masa molar = 15,999 u (oxygen)

R: ideal gas constant = 8.314472 cm^3 *MPa/K*mol =

grados K = °C + 273.15

P1*V*M/R*T = m1

P2*V*M/R*T = m2

masa released : m1-m2 = (P1-P2) * V*M/R*T

m2-m1 = 200 * 10^-3 MPa * 12 * 10^6 cm^3 * 15.999 u / 8.314472 (cm^3 * MPa/K *mol) * 290. 15 K

m2-m1= 38 397.6 * 10^3 u*mol / 2412.44 = 15916.5 g = 15.9165 kg

3 0
3 years ago
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