Top answers

Maths
All levels

The formula to convert from degrees Fahrenheit to degrees Celsius is C=(F-32)*(5/9). If it is 32 degrees Celsius, what is the temperature in Fahrenheit?

This question involves rearranging the equation to make F, the temperature in Fahrenheit, the subject of the formula. We'll start by writing out the original formula, and then we will move the elements of...

SB
Answered by Sarah B. Maths tutor
9971 Views

i) Factorise x^2 – 7x + 12 ii) Solve x^2 – 7x + 12

i)STEP 1: Make 2 brackets and put an x in each one: (x )(x )
STEP 2: Find 2 numbers when multiplied together make 12 and when added together they make -7
Multiples of 12:
1 x ...

HC
Answered by Hashir C. Maths tutor
3433 Views

Please expand the brackets in the following equation to get a quadratic equation. Then, please show using the quadratic formula that the solutions to the equation are x=3 and x=5. Here is the starting equation: (x-3)(x-5)=0

(X-3)(x-5)=0Use F.O.I.LFirst x multiplied by x gives x2outer x multiplied by -5 gives -5xinner x multiplied by -3 gives -3xlast -5x-3 gives +15combining we get x2-8x+15=0the quadrati...

JG
Answered by Jacob G. Maths tutor
3156 Views

Show that the funtion (x-3)(x^2+3x+1) has two stationary points and give the co-ordinates of these points

Stationary points are points where the gradient of a function is equal to 0. In this question the product rule can be used to find the gradient of the given function. The product rule is given by u'v+uv'...

JR
Answered by Joe R. Maths tutor
3038 Views

Find the nature of the turning points of the graph given by the equation x^4 +(8/3)*x^3 -2x^2 -8x +177 (6 marks)

(1 mark) Differentiate equation in the question: 4x3+8x2-4x-8(1 mark) Equate this to zero: (x-1)(x+1)(x+2)=0(1 mark) Find turning points (roots of above equation): x=1,-1,-2(1 mark) ...

EB
Answered by Elizabeth B. Maths tutor
3072 Views

We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences