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mr_godi [17]
3 years ago
14

10000 POINTS I NEED HELP THESE HAVE TO BE MADE AND PUT INTO FROM IMPOPER FRACTIONS TO MIXED NUMBERS I AM TIMED PLZ HELP I WILL G

IVE U BRANIY

Mathematics
1 answer:
kherson [118]3 years ago
8 0

Answer:

1)= 2 \frac{1}{3}

2) =1\frac{2}{5}

3)=2\frac{1}{2}

4)=2\frac{3}{10}

5)=2\frac{1}{6}

6)= 2\frac{5}{8}

7)= 1\frac{5}{8\\}

8)=2\frac{2}{3}

9)= 2  1/4

10)=  1  1/2

11) = 1 4/5

12)= 2  5/6

13)= 1 7/10

14)= 2 7/10

15)= 2 1/5

16)= 2 1/10

happy to help!!!

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The system of equations
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Answer:

z = -12

Step-by-step explanation:

The given system of equations is:

xy/(x + y) = 1 ...........................(1)

xz/(x + z) = 2...........................(2)

yz/(y + z) = 3...........................(3)

From (1): x + y = xy

=> y = xy - x

y = x(y - 1)

x = y/(y - 1).......................................(4)

From (2): 2(x + z) = xz

=> 2x + 2z = xz

2x = xz - 2z

2x = z(x - 2)

z = 2x/(x - 2) ....................................(5)

From (3): 3(y + z) = yz

=> 3y + 3z = yz

3y = yz - 3z

3y = z(y - 3)

z = 3y/(y - 3)....................................(6)

Comparing (5) and (6)

2x/(x - 2) = 3y/(y - 3)

2x(y - 3) = 3y(x - 2)

2xy - 6x = 3xy - 6y

6(y - x) = xy .................................(7)

But from (1): xy = x + y

Using this in (7), we have

6(y - x) = x + y

6y - y - 6x - x = 0

5y - 7x = 0

5y = 7x

x = 5y/7................................................(8)

Using this in (4)

5y/7 = y/(y - 1)

1/(y - 1) = 5/7

(y - 1) = 7/5

y = 1 + 7/5

y = 12/5..........................................(9)

Using this in (8)

x = 5(12/5)/7 = 12/7 .......................(10)

Using (10) in (5)

z = 2x/(x - 2)

z = 2(12/7) ÷ (12/7 - 2)

= 24/7 ÷ -2/7

= 24/7 × (-7/2)

= -24/2 = -12

z = -12.

4 0
4 years ago
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liubo4ka [24]
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3 years ago
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Step-by-step explanation:

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3 years ago
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Sloan [31]

Answer:

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4 0
3 years ago
Find all solutions of the equation in the interval [0, 2pi).
mina [271]

Answer:

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Step-by-step explanation:

-4 \sin x = 1 - cos^2 x

  • Rewrite it by using the identity \sin^2x + \cos^2x = 1

=> -4\sin x = \sin^2x

  • Add 4sin x to both the sides.

=> -4\sin x + 4\sin x = sin^2x + 4\sin x

=> \sin^2x + 4\sin x = 0

  • Take sin x common from the expression in L.H.S.

=> \sin x(\sin x + 4)=0

Here , we can get two more equations to find x.

1) \sin x(\sin x + 4)=0

  • Divide both the sides by sin x

=> \frac{\sin x(\sin x + 4)}{\sin x} = \frac{0}{\sin x}

=> \sin x + 4 = 0

  • Substract 4 from both the sides.

=> \sin x + 4 - 4 = 0 - 4

=> \sin x = -4

=> x = No \; Solution

2) \sin x(\sin x + 4)=0

  • Divide both the sides by (sin x + 4)

=> \frac{\sin x(\sin x + 4)}{\sin x + 4} = \frac{0}{\sin x + 4}

=> \sin x =  0

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5 0
3 years ago
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