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Kruka [31]
3 years ago
15

What number is the only whole number that is not a natural number?

Mathematics
2 answers:
slamgirl [31]3 years ago
7 0
The only whole number that isn’t a natural number is 0
lyudmila [28]3 years ago
5 0

Answer: 0

Step-by-step explanation: Natural numbers are the set of counting numbers that start at 0 an go on indefinitely.

Example ⇒ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20...}

When we include 0 in the set of natural numbers, we get the new set of numbers which is called the set of whole numbers.

Example ⇒ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20...}

All of the natural numbers are contained within the set of whole numbers.

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Find the height of a square pyramid that has a volume of 32 cubic feet and a base length of 4 feet. . . . a. 2 feet. b. 4 feet.
Mademuasel [1]
Volume of Pyramid = (1/3)*(base area) * height.

32 ft³  =  (1/3)*(4*4)* h.        Since Pyramid is square based.

32 = (16/3)*h

16h/3 = 32

h = 32*3/16

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4 0
3 years ago
HELP!!!! I think its C but I'm not sure!
MA_775_DIABLO [31]

Answer:

The fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are x=1\pm i\sqrt{7}.

Step-by-step explanation:

Consider the provided information.

Algebra's fundamental theorem states that: Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers.

Now consider the provided equation.

2x^2-4x+16=0

The degree of the polynomial equation is 2, therefore according to Algebra's fundamental theorem the equation have two complex roots.

Now find the root of the equation.

For the quadratic equation of the form ax^2+bx+c=0 the solutions are: x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Substitute a=2,\:b=-4,\:\ and \ c=16 in above formula.

x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\cdot \:16}}{2\cdot \:2}

x_{1,\:2}=\frac{4\pm \sqrt{16-128}}{4}

x_{1,\:2}=\frac{4\pm \sqrt{-112}}{4}

x_{1,\:2}=\frac{4\pm 4i\sqrt{7}}{4}

x_{1,\:2}=1\pm i\sqrt{7}

Hence, the fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are x=1\pm i\sqrt{7}.

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y=mx+b
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Hope this helps
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