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Verdich [7]
4 years ago
11

my dudes could you help me cause im really bad at math heres the problem choose all the values that make the inequality true. 28

-7x ≤ -4(-7x - 7) the options : -10 -5 0 5 10
Mathematics
1 answer:
Lubov Fominskaja [6]4 years ago
5 0

Answer:

  0, 5, 10

Step-by-step explanation:

You can solve the inequality to determine what values to select.

  28 -7x ≤ -4(-7x -7) . . . . . . given

  28 -7x ≤ 28x +28 . . . . . . eliminate parentheses*

  -7x ≤ 28x . . . . . . . . . . . . . subtract 28 from both sides

  0 ≤ 35x . . . . . . . . . . . . . . add 7x to both sides

  0 ≤ x . . . . . . . . . . . . . . . . . divide by 35

So, all values that are 0 or greater will make the inequality true:

  0, 5, 10

_____

* Parentheses are eliminated using the distributive property. The factor outside parentheses multiplies each of the terms inside. Pay attention to signs.

  -4(-7x -7) = (-4)(-7x) +(-4)(-7) = 28x +28

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Step-by-step explanation:

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5 0
3 years ago
I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu
ra1l [238]

The imaginary unit i belongs to the set of complex numbers, denoted by \mathbb C. These numbers take the form a+bi, where a,b are any real numbers.

The set of real numbers, \mathbb R, is a subset of \mathbb C, where each number in \mathbb R can be obtained by taking b=0 and letting a be any real number.

But any number in \mathbb C with non-zero imaginary part is not a real number. This includes i.

  • "is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because 2=2+0i.

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising i to the i-th power. Since i=e^{i\pi/2}, we have

i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079

3 0
3 years ago
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Answer:

Hello! Your answer would be below!

Step-by-step explanation:

Graph B)

Hope I helped! Ask me anything if you have questions!

3 0
3 years ago
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x=7

I'm sorry if I'm wrong

8 0
3 years ago
That is the correct answer
Arlecino [84]
3 and 4/8 for question number one
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