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3 years ago

The price of a gallon of heating oil rose from 1.60 a gallon to 1.92 . by what percent did the price increase?

Mathematics

2 answers:

Rudiy273 years ago

4 0

Answer: required percent increase in price=20%

Step-by-step explanation:

required percent increase in price= $\frac{1.92-1.60}{1.60}$ \times100

= $\frac{0.32}{1.60}$ \times100

= $0.20\times100$

=20%

AURORKA [14]3 years ago

3 0

Answer:

20%

Step-by-step explanation:

1.92 - 1.60 = 0.32

percentage increase = (0.32/1.60) * 100 =

0.2 * 100 = 20%

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Solve the equations and graph the solution

jenyasd209 [6]

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The inequalities are:

$-5x$ or $2x$

To find:

The solution for the given inequalities and graph the solution.

Solution:

We have,

$-5x$ or $2x$

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$-5x$

Divide both sides by -5.

$x>\dfrac{-3}{-5}$

$x>\dfrac{3}{5}$ ...(i)

And,

$2x$

Divide both sides by 2.

$x$

$x$ ...(ii)

From (i) and (ii). we get

$x$ or $x>\dfrac{3}{5}$

The interval notation of the solution is $(-\infty,-4)\cup \left(\dfrac{3}{5},\infty\right)$ .

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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