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# Tutor profile: Bronte F.

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Bronte F.
Senior Laboratory Technician
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## Questions

### Subject:Pre-Algebra

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Question:

Calculate: (-5)^3 + (2)^4

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Bronte F.

This problem is easy for the most part, if you are not familiar it would help to go over "Exponents with Negative bases". The key to this problem is watching your exponents and multiplication work, We will start with (-5)^3. Here is is asking to multiply -5 a total of 3 times together, so that means: -5 x -5 x -5 It is crucial to watch those negatives when you multiply so we will break it in parts: -5 x -5 = 25 25 x -5 = -125 Remember multiplying a negative and a negative gives you a positive! Then we focus on (2)^4: 2 x 2 x 2 x 2 = 16 But we are not done yet! Now we have to add them both together: -125 + 16 = -109

### Subject:Basic Math

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Question:

Find the product: -4x( -x^3+2x-5)

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Bronte F.

For this question it is asking for you to find the end product when solving, and to solve you will use the distribution rule. What you need to watch is when you multiply like terms together what that does to the exponent and signs. By distribution I am going to multiply the 4x to every term within the parenthesis. First I am going to multiply -4x and -x^3 which gives me 4x^4 Then -4x multiplied by 2x which equals -8x^2 Lastly -4x multiplied by -5 which equals 20x So your answer will be: 4x^4-8x^2+20

### Subject:Algebra

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Question:

Find the solutions of the equation: x^2 + 5x + 6 = 0

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Bronte F.

When an equation such as this one is set equal to 0 and it is asked "find the solutions of the equation", it is simply asking "what solutions to this equation make it zero?" or I could also say "what x values when plugged into the equation when worked out make it zero?" There are a few ways to solve this problem i.e factoring or by quadratic equation, but to save the headache and possibly the hate for fractions (like mine) I will solve this the easiest way I know how, by factoring. Factoring is a lesson in itself. For this problem I want to find a pair of factors for 6 when added together is 5. So that would be (x+2) and (x+3). x^2 + 5x + 6 = (x+2)(x+3) = 0 So now set your pair of factors equal to 0: x+2 = 0 x+3 = 0 Get like terms on one side: x = -2 x = -3 And those are your solutions!

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