Top answers

Maths
GCSE

How do I solve simultaneous equations?

A easy way to solve simultaneous equations is to substitiute one equation into another and then equate to find one of the unknowns. Once this has been found its a simple matter of substituting it back int...

KL
Answered by Kishen L. Maths tutor
3255 Views

Frank, Mary and Seth shared some sweets in the ratio 4:5:7. Seth got 18 more sweets than Frank. Work out the total number of sweets they shared.

The ratio 4:5:7 is a way of comparing the size of the shares received by Frank, Mary and Seth. It helps to pretend that we are sharing the sweets between some equally-sized boxes: each box has the same nu...

RN
Answered by Richard N. Maths tutor
28492 Views

Solve the simultaneous equations “x^2+y^2=4” and “x=2-y”. What does this tell us about the circle centred on the origin, with radius 2, and the straight line with y-intercept 2 and gradient -1?

In order to solve the pair of simultaneous equations, we must find a single set of values for x and y which fulfils both at once. By substituting “x=2-y” into “x2+y2=4”, we obtain a ...

AD
Answered by Aaron D. Maths tutor
11246 Views

What is the length of the hypotenuse in the right angled triangle to one decimal place? (a=5cm, b=4cm)

Step 1. Use the formula a2+b2=c2

Step 2. Square a and b to find a2=25 and b2=16

Step 3. Add a2 and b2 to find c

ER
Answered by Eleanor R. Maths tutor
3549 Views

Prove the quadratic formula for ax^2 + bx + c = 0, where a is non 0 and a,b and c are reals.

By completing the square: ax^2 + bx + c = 0 => x^2 + (bx)/a + c/a = 0 (divide both side by a, since a is non-zero) => (x + b/(2a))^2 + c...

SN
Answered by ShenZhen N. Maths tutor
8673 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