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A team of four is chosen from six married couples. If a husband and wife cannot both be on the team, in how many ways can the team be formed?

There are 12 choices for the first memebr, then 10 for the next (as it cannot be the spouse), then 8, then 6.Therefore there are 121086 permutations where partners aren't together.However, th...

Answered by Lily C. Maths tutor
2465 Views

How can I find the size of an angle in a right-angled triangle if I know the lengths of two of the sides?

We can find any angle in a right-angled (RA) triangle if we know the lengths of two of the sides, using SohCahToa.
The first step is to identify which of the sides we know. There are three sides in a...

Answered by Sarah M. Maths tutor
3231 Views

There are 700 students in a high school. 10% of them play team sports. 36 students play football, and 22 students play both football and basketball. When choosing one student from the school, what is the probability of them playing basketball only?

We know that 10% play team sports, that means 700*0.1=70 students play team sports. Out of those 70, 36 students play football, 22 play both sports, that means 36-22=14 play football only. Let's draw a di...

Answered by Dorottya T. Maths tutor
2570 Views

Calculate the binomial expansion of (2x+6)^5 up to x^3 where x is decreasing.

In order to use the binomial expansion, we must have an 'x' with no coefficients - so no number before it.
So we take out a factor of 2:(2(x+3))^5
We can then simplify to:32(x+3)^5
by expa...

Answered by Sophie C. Maths tutor
3425 Views

Find a solution for the following simultaneous equations: 2x+y=5 and 2x-5y=2

Subtract the 2nd equation from the first: 2x--2x+y+6y=5-2 <=> 7y=3 <=> y=3/7Input the value of y into one of the equations: 2x+3/7=5 <=> 2x=(35-3)/7 <=> x=16/7

Answered by Darshil C. Maths tutor
3079 Views

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