17772 questions

Find values of x in the interval 0<x<360 degrees. For which 5sin^2(x) + 5 sin(x) +4 cos^2(x)=0

This question is split up into two parts.
Firstly recall the trigonometric identities you know, the trick here is to eliminate one of the squared terms. Using 4sin^2(x) +4cos^2(x) = 4, the cos term is eliminated.
Rearranging this equation leaves you with a strange quadratic equation, but if you pretend sin is x it actually looks quite simple and can be solved like a simple quadratic. Solve like this and replace x for sin and the solution follows
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James G.

Answered by James, Maths tutor with MyTutor

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How do you know where the bonds link to form a new molecule?

Bonds in a molecule BREAK when that molecule is reacted with another molecule. They then MAKE another bond.This bond can be between molecules or can just be part of the new molecule.There are many types of bonds: single, double and triple, therefore depending which bonds are broken, different bond types can result.
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Nerea I.

Answered by Nerea, Chemistry tutor with MyTutor

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Let f be a function of a real variable into the real domain : f(x) = x^2 - 2*x + 1. Find the roots and the extremum of the function f.

Let f be a function of a real variable into the real domain : f(x) = x^2 - 2*x + 1. Find the roots and the extremum of the function f.
i) Root finding The function f is a 2nd degree polynomial. The root finding formula is hence applicable (Reminder if f(x) =a*x^2+b*x+c a 2nd degree polynomial, a, b and c real variables then its roots are defined by x = (-b +- sqrt(b^2-4ac))/(2*a) ) The determinant of the 2nd degree polynomial is delta = (-2)^2-4*1*1 = 0 . The function hence only has one repeated root given by x = (-(-2)+sqrt(0))/2*1 = 1 . ii) Finding the extremum To find the extremum of a function we need to analyse the behaviour of its 1st derivative. f is continuous in the real domain, its derivative is hence defined for all real x . f'(x) = 2*x - 2 f'(x) = 0 implies x = 1 . The extremum is hence located at x=1 and is the repeated root of the function. Before and after x=1, f(x) is strictly greater than 0, the extremum is hence a minimum. Those conclusions could have found by graphing the function.
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Matteo M.

Answered by Matteo, Maths tutor with MyTutor

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Effect of non-competitive inhibitor (on enzyme activity) on enzyme catalysed reaction

Binds on to enzyme at a site that is not the active sitechanges the tertiary structure of the active sitesubstrate cannot bindinhibits enzyme activity
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Kelvin C.

Answered by Kelvin, Biology tutor with MyTutor

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Effect of competitive inhibitor (on enzyme activity) on enzyme catalysed reaction

similar in shape to {substrate}fits onto active siteprevents substrate bindinginhibits enzyme activity
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Kelvin C.

Answered by Kelvin, Biology tutor with MyTutor

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Effect of enzyme concentration in excess substrate on enzyme catalysed reaction

directly proportional: increase in [enzyme] will increase rate of reaction
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Kelvin C.

Answered by Kelvin, Biology tutor with MyTutor

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Effect of substrate concentration on enzyme catalysed reaction

Rate of reaction increases proportional to increase in [substrate]Rate of reaction then levels off/ plateaus not all active sites are being used; adding substrate increases RoR all active sites used/ [enzyme] becomes limiting
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Kelvin C.

Answered by Kelvin, Biology tutor with MyTutor

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Effect of pH on enzyme catalysed reaction

At optimum pH, active site is complimentary to substratepH changes disrupt H-bonds/ ionic bonds
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Kelvin C.

Answered by Kelvin, Biology tutor with MyTutor

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