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A piece of wire is 240cm long. Peter cuts two 45 cm lengths off the wire. He then cuts the rest of the wire into as many 40cm length as possible. How many wires of 40cm can be cut?

Two 45cm lengths cut off 240cm long wire. 240 - (2*45) = 150cmHow many 40cm length wires can be cut out of the remainder 150cm wire?150/40= 3.75As we are looking for entire 40cm length wires, we disregard...

Answered by Federico B. Maths tutor
5920 Views

Find the area under the curve y=xsin(x), between the limits x=-pi/2 and x=pi/2.

We are going to need to use integration for this problem as it involves finding an area under a curve. Also notice that y is a product of two functions of x ; x and sin(x). This means in this particular c...

Answered by Oakley Y. Maths tutor
2277 Views

By integrating, find the area between the curve and x axis of y = x*exp(x) between x = 0 and x = 1

The area under a curve is found by integration.
Area = integral{xexp(x) dx} with limits 0 and 1
It's necessary to use integration by parts to find this integral as there are two x f...

Answered by Maths tutor
2346 Views

A shopkeeper compares the income from sales of a laptop in March and in April. The price in April was 1/5 more than in March. The number sold in April was 1/4 less than in March. By what fraction does income decrease from March to April?

Price in March = a Price in April = a x 1.2 = 1.2aNumber sold in March= b Number sold in April = b x 0.75 = 0.75bIncome = Price x Number soldIncome in March = abIncome in April = 0.9abDecrease is ...

Answered by Robert H. Maths tutor
7762 Views

Solve the simultaneous equations: 5x + y = 21, x - 3y = 9

5x + y = 21x - 3y = 9First, multiply the equation one by 3 so we can cancel out the y's.15x + 3y = 63 (we've now multiplied this by 3)x - 3y = 9Now we can add the two equations together, thus adding 3 and...

Answered by Sebastian P. Maths tutor
2439 Views

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