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Tamiku [17]
3 years ago
12

[tex]2yx^{4} x 5yx^{3}

Mathematics
1 answer:
Paladinen [302]3 years ago
4 0

Answer:B

Step-by-step explanation: hopes this helps

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Moesha needs $79 for a class trip. She already has $22 and she can earn the rest by babysitting for 8 hours. If h represents her
elena55 [62]

Answer:

79-22=8h

Step-by-step explanation:

She can earn the rest by babysitting for only 8 hours,

so take 22 from 79, divide the total by 8 and that is how much she gets per hour (8h)

6 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP 50 POINTS AND BRAINLYEST
Ludmilka [50]

Answer:

Either x = 3, y = 2

or

(3,2) is your answer.

Step-by-step explanation:

2x +y = 8  (1)

y = - x+ 5  (2)

Put equation (2) into equation (1). Use the y value in (2) to go into the y value in (1).

2x - x+5 = 8       Combine the x values

x + 5 =  8            Subtract 5 from both sides

x + 5-5 = 8-5

x=3

Now go back to (2). Put x in for the value of -x. Watch the sign.

y = -3+ 5

y = 2

So the answer is (3,2)

5 0
3 years ago
Solve the division problem. Round answer to the nearest hundreth<br><br> 9.2 divide 52.063
notsponge [240]
9.2/52.063 is equal to 0.18.
7 0
3 years ago
x2 + y2 − 4x + 12y − 20 = 0 (x − 6)2 + (y − 4)2 = 56 x2 + y2 + 6x − 8y − 10 = 0 (x − 2)2 + (y + 6)2 = 60 3x2 + 3y2 + 12x + 18y −
snow_tiger [21]
For this case, what we must do is fill squares in all the expressions until we find the correct result.
 We have then:
 
 x2 + y2 − 4x + 12y − 20 = 0 x2 + y2  − 4x + 12y = 20
 x2  − 4x + y2 + 12y = 20
 x2  − 4x + (12/2)^2 + y2 + 12y  + (-4/2)^2 = 20 + (12/2)^2 + (-4/2)^2
 x2  − 4x + (6)^2 + y2 + 12y  + (-2)^2 = 20 + (6)^2 + (-2)^2
 x2  − 4x + 36 + y2 + 12y  + 4 = 20 + 36 + 4
 (x − 2)2 + (y + 6)2 = 60 

 
3x2 + 3y2 + 12x + 18y − 15 = 0 
 
x2 + y2 + 4x + 6y − 5 = 0 
 x2 + y2 + 4x + 6y  = 5 
 x2  + 4x + (4/2)^2 + y2 + 6y + (6/2)^2 = 5 + (4/2)^2 + (6/2)^2 
 x2  + 4x + (2)^2 + y2 + 6y + (3)^2 = 5 + (2)^2 + (3)^2 
 x2  + 4x + 4 + y2 + 6y + 9 = 5 + 4 + 9 
 (x + 2)2 + (y + 3)2 = 18 

 2x2 + 2y2 − 24x − 16y − 8 = 0
 x2 + y2 − 12x − 8y − 4 = 0
 x2 + y2 − 12x − 8y = 4 
 x2 − 12x + (-12/2)^2 + y2 − 8y + (-8/2)^2 = 4 + (-12/2)^2 + (-8/2)^2
 x2 − 12x + (-6)^2 + y2 − 8y + (-4)^2 = 4 + (-6)^2 + (-4)^2
 x2 − 12x + 36 + y2 − 8y + 16 = 4 + 36 + 16
 (x − 6)2 + (y − 4)2 = 56 

 x2 + y2 + 2x − 12y − 9 = 0
 x2 + y2  + 2x - 12y = 9
 x2  + 2x + y2 - 12y = 9
 x2  + 2x + (2/2)^2 + y2 - 12y  + (-12/2)^2 = 9 + (2/2)^2 + (-12/2)^2
 x2  + 2x + (1)^2 + y2 - 12y  + (-6)^2 = 9 + (1)^2 + (-6)^2
 x2  + 2x + 1 + y2 - 12y  + 36 = 9 + 1 + 36
 (x + 1)2 + (y − 6)2 = 46
7 0
3 years ago
Based on a kc value of 0.200 and the given data table, what are the equilibrium concentrations of xy, x, and y, respectively?
MariettaO [177]
There is no given data table but based on the question, the reaction is
xy <=> x + y
If we let M as the initial concentration of xy and c as the in the concentration after the dissociation, then we can use the ICE method
    xy   <=>   x    +   y
I    M
C  -c            c         c
-----------------------------
E  M-c          c         c

Solve for c using
Kc = c(c) / (M - c)
And the concentration of the xy, x, and y can then be determined
3 0
3 years ago
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