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Mathematics

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

Associative.

Step-by-step explanation:

In addition and multiplication, it does not matter which two numbers you put together first. On the left aide 6 is associated with 3. On the right side 3 is grouped with 7. It doesn't matter where 3 goes. You still get the same answer.

(6 + 3) + 7 = 6 + (3 + 7)

9 + 7 = 6 + 10

16 = 16

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What is the common ratio of the sequence 1, 4, 16, 64...... <br> 1/3 <br> 4 <br> 3<br> 1/4

34kurt

Answer:

Explanation:

1, 4, 16, 64......

mutiply all of the numbers by 4

1*4=4

4*4= 16

16*4= 64

4 0

3 years ago

How do i solve this question?

Andrews [41]

when it comes to checking if a function is even or odd, it boils down to changing the argument, namely x = -x, and if the <u>resulting function is the same as the original</u>, then is even, if the <u>resulting function is the same as the original but negative</u>, is odd, if neither, well then neither :).

anyway, that said, let's first expand it and then plug in -x,

$\bf f(x)=(x^2-8)^2\implies f(x)=(x^2-8)(x^2-8)\implies f(x)=\stackrel{FOIL}{x^4-16x^2+64}\\\\[-0.35em]~\dotfill\\\\f(-x)=(-x)^4-16(-x)^2+64\qquad \begin{cases}(-x)(-x)(-x)(-x)=x^4\\(-x)(-x)=x^2\end{cases}\\\\\\f(-x)=x^4-16x^2+64\impliedby \stackrel{\textit{same as the original}}{Even}$

3 0

4 years ago

Find the solutions to the equation 102x 11 = (x 6)2 – 2. Which values are approximate solutions to the equation? Select two answ

otez555 [7]

You can try finding the roots of the given quadratic equation to get to the solution of the equation.

There are two solutions to the given quadratic equation

$x = 0.202, x = 113.798$

<h3>How to find the roots of a quadratic equation?</h3>

Suppose that the given quadratic equation is $ax^2 + bx +c = 0$

Then its roots are given as:

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

<h3>How to find the solution to the given equation?</h3>

First we will convert it in the aforesaid standard form.

$102x + 11 = (x-6)^2 - 2\\102x + 11 + 2 = x^2 + 36 - 12x\\0 = x^2 -114x + 23\\x^2 -114x + 23 = 0\\$

Thus, we have

a = 1. b = -114, c = 23

Using the formula for getting the roots of a quadratic equation,

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\x = \dfrac{114 \pm \sqrt{114^2 - 92}}{2} \\\\ x = 0.202 (\text{used "-" sign})\\\\x = 113.798 ( used "+" sign})$