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2 years ago

Which of the following is not a solution of the equation y=3x−4 ?

Mathematics

1 answer:

lisabon 2012 [21]2 years ago

6 0

Answer:

(-1, -1) is not a solution to the system

Step-by-step explanation:

Y = 3x - 4. Substitute the points for x and y

2 = 3(2) - 4

2 = 2 solution

y = 3x - 4

-4 = 3(0) - 4

-4 = -4 solution

y = 3x - 4

8 = 3(4) - 4

8 = 8 solution

y = 3x - 4

-1 = 3(-1) - 4

-1 = -7 not solution

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3 years ago

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1 year ago

1.The equation of line CD is (y−3) = − 2 (x − 4). What is the slope of a line perpendicular to line CD? 2.The equation of line A

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So here's how to do it,
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for perpendicular lines flip the slope

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3 years ago

Read 2 more answers

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