W^3=125 when w = ...........

Mathematics

1 answer:

blondinia [14]2 years ago

5 0

Answer: w = 5

Step-by-step explanation:

Step 1: Take cube root.

w=(125) ^(1/3)

W = 5

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Given a function <img src="https://tex.z-dn.net/?f=f%28x%29%3D3x%5E4-5x%5E2%2B2x-3" id="TexFormula1" title="f(x)=3x^4-5x^2+2x-3"

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$\huge\boxed{f(-1) = -7}$

Step-by-step explanation:

In order to solve for this function, we need to substitute in our value of x inside to find f(x). Since we are trying to evalue f(-1), we will substitute -1 in as x to our equation.

$f(-1) = 3(-1)^4 - 5(-1)^2 + 2(-1) - 3$

Now we can solve for the function by multiplying/subtracting/adding our known values.

Starting with the first term to the last term:

$3(-1)^4 = 3$

WAIT! How is this possible? $-1^4 = -1$ (according to my calculator), and $3 \cdot -1 = -3$ , not 3!

It's important to note that taking a power of a negative number and multiplying a negative number are two different things. Let's use $-2^2$ as an example.

What your calculator did was follow BEMDAS since it wasn't explicitly told not to.

BEMDAS:

- Brackets

- Exponents

- Multiplication/Division

- Addition/Subtraction

Examining the equation, your calculator used this rule properly. Note that exponents come over multiplication.

So rather than being "-2 squared" - it's "the negative of of 2 squared."

Tying this back into our problem, the squared method would only be true if it looks like $-1^4$ . However, since we're substituting in -1, it looks like $(-1)^4$ , so the expression reads out as "-1 to the fourth."