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monitta
3 years ago
9

Decompose the fraction 7/9 = show work

Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
7 0
Hey You!

7/9 = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9

7/9 = 1/9 + 6/9

7/9 = 4/9 + 3/9

7/9 = 5/9 + 2/9

That's it! =)
Kamila [148]3 years ago
4 0
A fraction can be expressed as the product of a whole number and a unit fraction....
So...

7 * 1/9
\frac{1}{9}  + \frac{1}{9}  + \frac{1}{9}  + \frac{1}{9}  + \frac{1}{9}  + \frac{1}{9}  + \frac{1}{9}

Start with a numerator of 1 and go up to 6.

1. \frac{1}{9} +  \frac{6}{9}
2. \frac{2}{9} +  \frac{5}{9}
3. \frac{3}{9} +  \frac{4}{9}
4. \frac{4}{9} +  \frac{3}{9}
5. \frac{5}{9} +  \frac{2}{9}
6. \frac{6}{9} +  \frac{1}{9}

Hope this helps!
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E=300-10 (30)<br> How to solve it step by step
Zepler [3.9K]

Answer:

Step-by-step explanation:

e = 300-10(30)

e = 300 - 300

e = 0

6 0
4 years ago
If the signs of the coordinates of collinear points P(-6,-2), Q(-5,2), and R (-4,6) reversed, are the 3 new points still colline
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If all the points have the signs reversed, therefore this means that the whole line is reflected about an axis. Since all are reflected therefore this further means that the new points are still collinear.

 

We can prove this by plotting the points:

P’(6,2), Q(5,-2), R(4,-6)

<span>From the graph, they are in 1 line hence collinear.</span>

6 0
3 years ago
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pochemuha
86 minutes would have passed.
8 0
3 years ago
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Does anyone know the answer to this one ?
DiKsa [7]

Answer:

(1, -4), (4, -1)

Step-by-step explanation:

A solution to a system of equations is when the two lines intersect. In this case, we have two intersections, which means two solutions.

From left to right, we first see an intersection at (1, -4).

Continuing to the right, we see another intersection at (4, -1)

These are our two solutions.

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Hope this helps!

4 0
3 years ago
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Simplify: (sin θ − cos θ)^2 − (sin θ + cos θ)^2 A) −4sin(θ)cos(θ) B) 2 C) sin^2 θ D) cos^2 θ
N76 [4]

Answer:

-4sinθcosθ

Step-by-step explanation:

Note:

1. (a + b)^2 = a^2 + 2ab + b^2

2. (a - b)^2 = a^2 - 2ab + b^2

3. sin^2θ + cos^2θ = 1

(sinθ -cosθ)^2 - (sinθ + cosθ)^2

= sin^2θ - 2sinθcosθ + cos^2θ - (sin^2θ + 2sinθcosθ + cos^2θ)

= sin^2θ + cos^2θ - 2sinθcosθ - (sin^2θ + cos^2θ + 2sinθcosθ)

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= -4sinθcosθ

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4 years ago
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