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3 years ago

PLEASE HELLLLLLLPPPPPP!!!!!

Mathematics

1 answer:

MariettaO [177]3 years ago

8 0

Answer:

Step-by-step explanation:

First, let's define the types of numbers.

Real numbers: a value that represents a quantity along a continuous line

ex: 1, 5, -6, 0.125, 45.5258,

Integer: Zero, any positive or negative number

ex: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5....

Whole numbers: a number without fractions [not negative]

ex: 0, 1, 2, 3, 4, 5.....

Rational numbers: any number that can be expressed as the quotient or fraction

ex: 1, 2, 3, 4, 5

Irrational numbers: any number that can't be expressed at the quotient or fraction

ex: pi, [3.14.... (never ending decimal)] square root of 2....

Based on this information, the false statment would be A. I believe, because

pi is a real number and it is an irrational number.

I hope this helps!!! :)

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Step-by-step explanation:

The given equation is:

$0=(3x+2)(x-4)$

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

$(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}$

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply $(3x+2)(x-4)$ i.e. $3x^2-12x+2x-8$ = $3x^2-10x-8$

So, the equation becomes:

$0=3x^2-10x-8$

Comparing this equation to standard quadratic equation:

$ax^2+bx+c=0$

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}$