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romanna [79]
3 years ago
6

Dennis made a scale drawing of his backyard using the scale 1/4 inch =3 feet the rectangular swimming pool was 2 inches long and

1 inch wide the drawing what was.the area of actual swimming pool
Mathematics
1 answer:
seraphim [82]3 years ago
3 0
(1/4)x = 2 

x = 8 x 3 = 24 feet long

(1/4)x = 1

x = 4 x 3 = 12 feet wide

24ft x 12ft = 288ft^2
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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
Tan x + sqrt(3) = - 2 tan x
Svet_ta [14]

9514 1404 393

Answer:

  nπ -π/6 . . . for any integer n

Step-by-step explanation:

  tan(x) +√3 = -2tan(x) . . . . . given

  3tan(x) = -√3 . . . . . . . . . . . add 2tan(x)-√3

  tan(x) = -√3/3 . . . . . . . . . . divide by 3

  x = arctan(-√3/3) = -π/6 . . . . use the inverse tangent function to find x

This is the value in the range (-π/2, π/2). The tangent function repeats with period π, so the set of values of x that will satisfy this equation is ...

  x = n·π -π/6 . . . . for any integer n

8 0
2 years ago
What is the inequality for 14-3x<-1
Maksim231197 [3]

Answer:

x > 5

Step-by-step explanation:

14 - 3x < -1

-3x < -15

x > 5

5 0
3 years ago
What is the area of a rectangle with vertices at (6,-3), (3,-6), (-1,-2), and (2,1). Enter your answer In the box
vladimir2022 [97]

Answer:

24 units

Step-by-step explanation:

Start by plotting on the rectangular coordinate system the four points given: (6,-3), (3,-6), (-1,-2) , and (2, 1)

Please see attached image for guidance with the answer.

These four points are represented with green dots in the image.

Notice that such rectangle can be viewed as inscribed in a larger square that is 7 units by 7 units, and on which these four points (the vertices of the rectangle) touch the square's perimeter.

So to make the calculation simple, we can find the area of the larger square (very simple to calculate since it is 7 times 7 = 49) and subtract from it the area of the four triangles that are depicted in yellow and in light green. Such result will give us the area of the rectangle (white background).

Out of the four triangles whose areas we need to extract, we notice that they are all right angle triangles with simple formulas to calculate their areas: base times height divided by 2.

The triangles in green have base 3 and height 3 so their areas are: (3 * 3) /2 = 4.5. since they are two of these the total area of the green triangles is 4.5*2 = 9

We do something similar with the yellow triangles, noticing that their base and heights are 4 and 4 respectively which gives an area of (4*4)/2 = 8

Again, since there are two of them, we multiply this by 2: 8 * 2 = 16

Now to calculate the area of the white rectangle we do:

Area of big square minus area of green triangles minus area of yellow triangles: 49 - 9 - 16 = 24

3 0
3 years ago
The weight of a nickel is 5 g. how many mg does a nickel weigh?
alexandr1967 [171]
There are 1000 milligrams in a gram (milli: 1/1000). Multiply 1000 by 5 g.

Final answer: 5000 mg.
4 0
3 years ago
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