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Grace A.
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Pre-Algebra
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Question:

Reduce the equation: (5+25)*10/5-160+3^2

Grace A.

1.) First we need to remember PEMDAS: Parenthesis Exponents Multiplication Division Addition Subtraction This is the order of operations. We must start with Parenthesis and end with Subtraction. Another note is that multiplication and division are equal in order. Subtraction and addition are also equal in order. The way we decide which one to do first is to start from left to right. 2.) Parenthesis: We must look for parenthesis. We have parenthesis around 5+25, so we must add them together before doing anything else: (30)*10/5-160+3^2 = 30*10/5-160+3^2 3.) Exponents: Now we look for any exponents in the equation. We see 3^2, so now we can find the result of 3^2. 3^2 = 3*3 = 9 Plug this back in: 30*10/5-160+9 4.) Multiplication/Division: We begin from the left side and look for any multiplying or dividing. The first thing we come across is: 30*10. 30*10 = 300 Now we can plug this back in: 300/5-160+9 Now look for more division/multiplication. We see 300/5. 300/5 = 60 Plug this back in: 60-160+9 5.) Subtraction/Addition We again begin from the left side and move towards the right. We first see 60-160. 60-160 = -100 (To subtract using a larger second number you can pretend it is 160-60=100, but then remember that 160 is negative, so the answer must also be negative.) Plug this in: -100+9 = -91 (You can rearrange this to read 9-100. Then you can think of it as 100-9=91. Because 100 > 9 and 100 is negative you should make the answer negative as well.) 6.) The answer is: -91

Calculus
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Question:

Find dy/dx y=3x^2/4x+x^2

Grace A.

1.) To find the derivative we must use the rule: [(Denominator*Derivative of numerator)-(Numerator*Derivative of Denominator)]/Denominator^2 2.) A good way to remember this rule is to think: Low Di-High minus High Di-Low Over Low Squared 3.) The first thing we need to do is separate the numerator and denominator: High = 3x^2 Low = 4x+x^2 4.) Take the derivatives of the numerator and denominator: Di-High = 6x Di-Low = 4+2x 5.) Then apply the formula: dy/dx = [(4x+x^2)*(6x)-(3x^2)(4+2x)]/(4x+x^2)^2 6.) Now all we have to do is reduce the answer. First Multiply (4x+x^2)(6x) (4x+x^2)(6x) = 24x^2+6x^3 Plug it back into the equation: dy/dx = [(24x^2+6x^3)-(3x^2)(4+2x)]/(4x+x^2)^2 7.) Then we can multiply (3x^2)(4+2x) (3x^2)(4+2x) = 12x^2+6x^3 Plug it back into the equation: dy/dx = [(24x^2+6x^3)-(12x^2+6x^3)]/(4x+x^2)^2 8.) Now it is best to distribute the minus sign in the numerator: dy/dx = (24x^2+6x^3-12x^2-6x^3)/(4x+x^2)^2 9.) Now we can cancel out some of the terms on the numerator: dy/dx = (12x^2)/(4x+x^2)^2 10.) Now the numerator is reduced we can multiply out the denominator: (4x+x^2)^2 = (4x+x^2)(4x+x^2) = 16x^2+4x^3+4x^3+x^4 = 16x^2+8x^3+x^4 = x^2(16+8x+x^2) Now plug it back into the equation: dy/dx = (12x^2)/[x^2(16+8x+x^2)] 11.) Now we can reduce the equation by removing the x^2 from the numerator and denominator. dy/dx = (12)/[(16+8x+x^2)] 12.) The finished derivative is: dy/dx = (12)/[(16+8x+x^2)]

Algebra
TutorMe
Question:

Solve this equation for X: (5*X)/2+5=25

Grace A.

1.) The goal is to move everything except X to the right side of the equal sign. 2.) The first thing we move is +5 This gives us: (5*X)/2=25-5 3.) This can reduce to: (5*X)/2=20 4.) Now that the right side is reduced fully we can look at the left side again. The next step is to move the /2 to the right side. The way we cancel out the division is by multiplication. This leaves us with: (5*X)=20*2 (5*X)=40 5.) Now we can get rid of the parenthesis around (5*X) because there are no other numbers or operations on the left side of the equation. That leaves us with: 5*X=40 6.) Now we can move the 5* to the right side of the equation. The way we move it over is by division. This is because division cancels out multiplication. That gives us: X=40/5 X=8 7.) Now we have the solution: X=8

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