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MArishka [77]
3 years ago
11

What is 2 7/9 as a improper fraction

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
5 0

multiply the denominator by the whole number, 2x9=18

then add the numerator the your answer, 18+7=25

but the denominator stays the same, hence your answer, 25/9

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Rosa bought 1 pound of cashews and 2 pounds of peanuts for $10. At the same store, Sabrina bought 2 pounds of cashews and 1 poun
Valentin [98]

Answer:

The solutions are x=4, y=3. Cashews are $4 per pound and Peanuts are $3 per pound.

Step-by-step explanation:

let x = cashews

let y = peanuts

x + 2y = 10 - rosa's purchase

2x + y = 11 - sabrina's purchase

subtract 2x to isolate 'y'

y = 11 - 2x

x + 2(11 - 2x) = 10

x + 22 - 4x = 10 (Use the distributive property.)

-3x = -12

x = 4.00

Plug this into the original equations to find "y"

2(4) + y = 11

8 + y = 11

y = 3

4 + 2y = 10

2y = 6

y = 3

the solutions are: x = 4. y = 3.

5 0
3 years ago
Gracie is building a square sandbox with sides 2 feet long. She wants to put sand 1.55 feet deep in the box. How much sand shoul
ASHA 777 [7]
2*2*1.55=6.2 cubic feet of sand
6 0
3 years ago
Read 2 more answers
Dorothy Kaatz , programmer regular hourly rate of $9.15. Double time for overtime.what is her straight time pay. What is her ove
Pepsi [2]

Answer:

regular pay per hour =9.15

overtime pay per hour =13.725

4 0
3 years ago
Slopes 3 and (1,4) is on the line put in standard form
VARVARA [1.3K]
Y - y₁ = m(x - x₁)       slope(m) = 3     point (1, 4)

y - 4 = 3(x - 1)

y - 4 = 3x - 3

y = 3x - 3 + 4

3x - y = -1  <<< your answer

hope this helps, God bless!
5 0
3 years ago
Find an equation of the plane. The plane that passes through the line of intersection of the planes x − z = 3 and y + 2z = 3 and
IgorLugansk [536]

Answer:

x-y-3=0

Step-by-step explanation:

Choose two common points on the line of intersection of the planes x − z = 3 and y + 2z = 3:

  • 1st point: if x=0, then z=-3 and y=3-2z=3+6=9, so the 1st point is (0,-3,9);
  • 2nd point: if x=3, then z=0 and y=3-2z=3-0=3, so the 2nd point is (3,0,3).

The perpendicular plane to the plane x+y-4z=6 is parallel to the vector with coordinates (1,1,-4) (normal vector of the given plane).

Hence, the equation of needed plane is

\left\|\begin{array}{ccc}x-0&y-(-3)&z-9\\3-0&0-(-3)&3-9\\1&1&-4\end{array}\right\|=0\Rightarrow \left\|\begin{array}{ccc}x&y+3&z-9\\3&3&-6\\1&1&-4\end{array}\right\|=0

Thus,

\left\|\begin{array}{ccc}x&y+3&z-9\\3&3&-6\\1&1&-4\end{array}\right\|=0\Rightarrow \\\\-12x-6(y+3)+3(z-9)-3(z-9)+6x+12(y+3)=0\\ \\-6x+6(y+3)=0\\ \\-x+y+3=0\\ \\x-y-3=0

4 0
4 years ago
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