Mathematics For Engineering Technicians Level 3 Answers

It is abit unclear but I am doing unit 4: mathematics for engineering technicians. I am stuck on both tasks, someone please help

Can even read it but I do a level maths so I can help if you like

way too fuzzy to see...anyway the maths involved in the level 3 BTEC in engineering is incredibly easy...it's essentially a watered down version of A-level maths...there's tons of video tutorials on YouTube for that, check em out.

(Original post by Froppy)
way too fuzzy to see...anyway the maths involved in the level 3 BTEC in engineering is incredibly easy...it's essentially a watered down version of A-level maths...there's tons of video tutorials on YouTube for that, check em out.

Lol bit harsh but true

Sorry it was a bit now i look back >_< was just talking frankly from my own personal experience with the course. When i did it there was a number of people who were just lazy and would ask others to do their maths for them...i kind of got that vibe from this thread.

(Original post by JordanStewart)
Can even read it but I do a level maths so I can help if you like

Sorry, I will send you a picture of task 2 which I'm struggling with atm

believe, I was also going to take engineering as an A level. but I would have to start year 12 again for me to do it. so i didnt do it because i only have few months until I start university and what would be the point to restart sixth form if i am already doing good. so i just hope that the grades i get this year helps me progress on to software engineering at university.

(Original post by redxreviver)
Sorry, I will send you a picture of task 2 which I'm struggling with atm

So it wants you to draw a graph with voltage on the x axis and power on the y.
To find the gradient. You need to use change in y divided by change in x. So draw your line of best fit. The pick two points on the line and you can work out the gradient.

To find the y intercept it's where the line crosses the y axis. So should be easy enough to read of the graph.

Y= mx+c. Well m is the gradient which you've worked out and c is the y intercept so you can put those values into the correct form and that's that.

Do power is the y value and voltage is x. So set x = 36 (I think can't quite make out the number) and put that into your y=mx+c equation and that will give you the answer.

(Original post by Snoozinghamster)
So it wants you to draw a graph with voltage on the x axis and power on the y.
To find the gradient. You need to use change in y divided by change in x. So draw your line of best fit. The pick two points on the line and you can work out the gradient.

To find the y intercept it's where the line crosses the y axis. So should be easy enough to read of the graph.

Y= mx+c. Well m is the gradient which you've worked out and c is the y intercept so you can put those values into the correct form and that's that.

Do power is the y value and voltage is x. So set x = 36 (I think can't quite make out the number) and put that into your y=mx+c equation and that will give you the answer.

Thank you, I have just done it