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1 year ago

Answer the following equations:

Mathematics

1 answer:

Andrew [12]1 year ago

7 0

Step-by-step explanation:

3(t+5) = 9

3t+15 = 9

3t = -6

t = -2

2(f-7) = -10

2f - 14 = -10

2f = 4

f = 2

-(c - 9) = 4

-c + 9 = 4

-c = -5

c = 5

-6(2t + 8) = -84

-12t - 48 = -84

-12t = -36

12t = 36

t = 3

-10 (s + 2) = -57

-10s - 20 = -57

-10s = -37

s = 3.7

7(3w + 8)/3 = -9

By cross multiplication,

7(3w + 8) = -27

21w + 56 = -27

21w = -83

w = -3.95

35/5 = (F - 32)/9

7 = (F - 32)/9

By cross multiplication,

63 = F - 32

63 + 32 = F

95 = F

Hope it helps.

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2 years ago

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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.

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"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