Is there a picture of the problem you can add to this?
Answer:
A. 0
General Formulas and Concepts:
<u>Pre-Algebra
</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I
</u>
<u>Pre-Calculus
</u>
<u>Calculus
</u>
- Derivatives
- Derivative Notation
- Derivative of csc(x) =
![\frac{d}{dx} [csc(x)] = -csc(x)cot(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bcsc%28x%29%5D%20%3D%20-csc%28x%29cot%28x%29)
Step-by-step explanation:
<u>Step 1: Define</u>
<u />
<u />
<u />
<u>Step 2: Differentiate</u>
- Differentiate:
![\frac{d}{dx} [csc(x)] = -csc(x)cot(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bcsc%28x%29%5D%20%3D%20-csc%28x%29cot%28x%29)
<u>Step 3: Evaluate</u>
- Substitute in <em>x</em>:

- Evaluate:

Answer:
no
Step-by-step explanation:
It does.
5x-2=0
5x=2
x=2/5
(2/5, 0)
is an x-intercept
Hello! I can help you with this!
a. The function that would best represent Samantha's account is f(5) = 500(1 + 0.04)^5. This is because $500 is the principal, the interest rate is 4%, and we're looking for the amount in the savings account 5 years later.
b. Okay. 1 = 0.04 is 1.04. 1.04^5 is 1.216652902. It's a long decimal, but don't delete it. Multiply that decimal by 500 and you get 608.32645 and other numbers behind it or 608 when rounded to the nearest dollar. Samantha will have about $608 in her savings account in 5 years.
Note: The formula goes like this: f(x) = P(1 + r)^x. This means, you add 1 and the simple interest rate in decimal form together and raise that up by the exponent. There is no shortcuts for this, so you'll have to use the calculator. There will be a very long decimal, but don't clear it. Instead, multiply it by the principal to get the answer. It seems very complicating, but if you do this right, it gets easier overtime and you'll make less errors. There are more complex problems out there, so this formula is very important, but it was kept simple for this question.
Answer:
None of these.
Step-by-step explanation:
Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.
Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.
In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.
If we look at the first term x^3 and we divide it by 13 we get
we cannot reduce it so it is not a fraction so 13 is not a factor of P
None of these is the right option.