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zvonat [6]
3 years ago
15

Helppppppppppppppppppppppppp[

Mathematics
1 answer:
RoseWind [281]3 years ago
3 0
Given two functions g (x) and h (x), it is called a function composed of g with h, and we write g o h:
 We have:
 g [h (x)]
 We have then:
 g [h (x)] = ((4-x) +1) / ((4-x) -2)
 We rewrite:
 g [h (x)] = (5-x) / (2-x)
 then:
 g [h (-3)] = (5 - (- 3)) / (2 - (- 3))
 g [h (-3)] = 8/5
 Answer:
 8/5
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What is the perimeter of rhombus LMNO?
kicyunya [14]

If you see that 3x-3 and set it equal to x+7 then you will get this,

3x-3=x+7


Now solve for X. You will get 5.

Then you plug five in and get 12 For both sides and then multiply 12 by 4 and get 48.

You answer is 48.

Hope this helps you.

4 0
3 years ago
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Could I have the answer to factorise 16a+12b
Tanzania [10]
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3 years ago
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- 5x - 5 = 3x + 19 what would X be?
Tanzania [10]

-5x - 5 = 3x + 19

---Move the x's to one side

-5x - 3x - 5 = 3x - 3x + 19

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---Isolate the -8x by removing the -5 from the left side

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---Divide both sides by -8 to get x by itself

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2 years ago
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If z=5-2i then |z| is equal to what ?
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<span>z=5-2i

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5 0
3 years ago
A positive real number is 6 less than another. If the sum of the square of the two numbers is 38, then finds the numbers.
Serggg [28]

Answer:

There seems to be a typo error but still this has solution.

Let the number be = x

A positive real number is 6 less than another. Means the second will be (x+6)

The sum of the square of the two numbers is 38.

x^{2} +(x+6)^{2} =38  

=> x^{2} +36+12x+x^{2} =38

=> 2x^{2}+12x+36 =38

=> 2x^{2}+12x=2

=> 2x^{2}+12x-2=0

Taking out 2 common;

=> x^{2}+6x-1=0

Solving this quadratic equation, we get;

x=-3+\sqrt{10} and x=-3-\sqrt{10}

As positive number is needed, we have -3+\sqrt{10}

or x = \sqrt{10}-3

x = 0.162277

And other number is 6.162277.

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3 years ago
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