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Nezavi [6.7K]
3 years ago
5

Factorise

y%7D" id="TexFormula1" title=" \sf{ {x}^{2} - {y}^{2} - x + y}" alt=" \sf{ {x}^{2} - {y}^{2} - x + y}" align="absmiddle" class="latex-formula">
Show workings too! ​
Mathematics
2 answers:
dem82 [27]3 years ago
8 0

Answer:

(x-y) (x+y-1)

Step-by-step explanation:

x2-y2-x+y

= (x-y) (x+y) - (x-y)

= (x-y) (x+y-1)

9966 [12]3 years ago
6 0

Answer:

(x-y) (x+y-1)

Step-by-step explanation:

x^2-y^2-x+y

(x-y) (x+y) - (x-y)

(x-y) (x+y-1)

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Find the GCF<br> 34 &amp; 68
photoshop1234 [79]

Answer:

The GCF would be 34.

34/34=1

68/34=2

If this answer was helpful please consider giving brainliest!

4 0
3 years ago
It takes an airplane 5/12 of a minute to fly 3/4 of a mile. What is the airplanes speed per minute?
Nikitich [7]
To make this a little easier, you would want to turn 3/4 into 9/12, to match 5/12's denominator.

You would then just add 7/1 to both fractions

(5/12+7/1=12/12 or 1 minutes)
(9/12+7/1=16/12=1 1/3)

So your answer would be 1.333... miles per minute

Hope this helps
7 0
3 years ago
Help needed now with Math question.
saveliy_v [14]

Answer:

is the last one

Step-by-step explanation:

5 0
3 years ago
Plz help will choose brainliest provide an explanation.
nika2105 [10]

Answer:

I wanna say b

Step-by-step explanation:

I'm gonna say b because u would think since they've paid $13, and it sells for $22, u would add that on the price, which in my opinion would be x + 13 = 22

7 0
3 years ago
Two cables support a 800​-lb ​weight, as shown. Find the tension in each cable.
jeka94

Answer:

  • 892 lb (right)
  • 653 lb (left)

Step-by-step explanation:

The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...

  Rcos(45°) +Lcos(75°) = 800

  Rsin(45°) -Lsin(75°) = 0

Solving these equations by Cramer's Rule, we get ...

  R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))

     = 800sin(75°)/sin(120°) ≈ 892 . . . pounds

  L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds

The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.

_____

This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...

  (right, left) = w/sin(α+β)×(sin(β), sin(α))

5 0
3 years ago
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