Name number that is rational but not and interger or a whole number

1 answer:

iogann1982 [59]3 years ago

3 0

Providing a short wikipedia definition here: "In mathematics, a rational number is any number that can be expressed as the quotient or fraction"

So any fraction would work for this question I believe?

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Convert 200 centimeters into kilometers

Temka [501]

200 centimetes converted to kilometers = 0.002

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4 years ago

The product of two consecutive numbers is 182

kupik [55]

2 consecutive numbers....x and x + 1

x(x + 1) = 182
x^2 + x = 182
x^2 + x - 182 = 0
(x - 13)(x + 14) = 0

x - 13 = 0
x = 13

x + 14 = 0
x = -14....not this one because it is negative

x + 1 = 13 + 1 = 14

ur 2 numbers are : 13 and 14

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3 years ago

What is the interval in which this function is positive?

NeX [460]

Answer:

(-∞,∞)

Step-by-step explanation:

It is positive for all x

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3 years ago

Solve 3x +5 =14 please help me

alisha [4.7K]

3x=14-5
3x=14-5
3x=9
x=9 ➗ 3
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the answer is 3

4 0

4 years ago

Read 2 more answers

The function f(x) = 2x2 + 3x + 5, when evaluated, gives a value of 19. What is the function’s input value?

goldenfox [79]

$f(x)=\quad 2{ x }^{ 2 }+3x+5$

Let's called the input 'z'

When we plug 'z' in the function we get ;

$f(z)=\quad 2{ z }^{ 2 }+3z+5$

And we know that, this is equal to 19, so ;

$2{ z }^{ 2 }+3z+5=\quad 19$

Let's rearrange the equation.

$2{ z }^{ 2 }+3z+5=\quad 19\\ \\ 2{ z }^{ 2 }+3z=\quad 19-5\\ \\ 2{ z }^{ 2 }+3z=\quad 14\\ \\ 2{ z }^{ 2 }+3z-14=\quad 0$

So we have a quadratic equation here.

We'll use this formula to solve it :

$\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

The formula is used in equation formed like this :

$a{ x }^{ 2 }+bx+c=0$

In our equation,

a=2 , b=3 and c=-14

Let's plug in the values in the formula to solve,

$a=2\quad b=3\quad c=-14\\ \\ \frac { -3\pm \sqrt { 9-(4\cdot 2\cdot -14) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9-(-112) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9+112 } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 121 } }{ 4 } \\ \\ \frac { -3\pm 11 }{ 4 }$

So,

$z=\quad \frac { -3+11 }{ 4 } \quad ,\quad \frac { -3-11 }{ 4 } \\ \\ z=\quad \frac { 8 }{ 4 } \quad ,\quad \frac { -14 }{ 4 } \\ \\ z=\quad 2,\quad -\frac { 7 }{ 2 }$

So the input can be both, 2 and $-\frac{7}{2}$

8 0

4 years ago