Recent content by ai93

Having trouble doing bending moments and points of contraflexture. That's why I need help!

Hi I am attempting a question on moments and beams. I have attached the question and my solutions so far. But having trouble completing d) and e) Can someone show how to work out bending moments and the point of contraflexure? [PLAIN]http:// any help is appreciated thanks

Is this the formula? What values would you need to sub in?

a) Find the roots of the equation $$x^{2}+5x-6$$ b) Sketch the graph of the function $$x^{2}+5x-6$$ labeling the points at which the graph crosses the axes and the co-ordinates of the maximum and minimum of the curve c) Find the equation of the tangent at the point where $$x=2$$ on the curve...

Beginning to understand this a lot easier. Just a question on the last part. How did you rearrange to get $$\theta$$? It went from, $$A=\frac{1}{2}r^2\left(\frac{s}{r}\right)$$ which I understand to $$=\frac{rs}{2}$$?

Thanks! $$2x+11=x+8$$ $$2x-x=8-11$$ $$\therefore x=-3$$ sub $$x=-3 into y=2x+11$$ = y=5 for c) Why is finding the distance necessary? Since we have to use the gradient/slope formula? Nevertheless m=$$\frac{5-3}{(-3)-(-4)}$$ m=2 :D

$$\tan\left(25^{\circ}\right)=\frac{w}{120}\tag{1}$$ $$w=tan(25)x120=55.95$$ $$tan(30)=\frac{55.95+h}{120}$$ $$\therefore tan(30) \cdot 120 = 55.95+h$$ $$69.28+55.95+h$$ $$69.28-55.95=h$$ $$\therefore h=13.33$$ I think that's right - - - Updated - - - If I am given a similar question...

Thanks this really did help :D

I have an equation of a line question a) Find the equation of the straight line with gradient 2 passing through point A (-4,3) I worked out the equation of the line, which is, y=2x+11. But having trouble with question b) and c) b) if the line in part a) intersects the line y=x+8 at point B...

Im finding trouble understanding the question. "Form two equations involving r and $$\theta$$" Basically means transpose the formula or make the equation equal to r and $$\theta$$? Since the arc length = $$s=r\theta$$ (I thought arc length was length=$$\frac{n}{360}\cdot2\pi(r)$$...

A vertical flagpole is fixed at the top of a vertical wall. From a point which is 120m measured horizontally from the base of the wall the angle of elevation to the top of the flagpole is 30 degrees, and the angle of elevation to the bottom of the flagpole is 25 degrees. a) Draw a clearly...

The length of the minor arc of a circle is 10cm, while the area of the sector AOB is 150cm2. a) Form two equations involving r and θ, where θ is measured in radians. b) Solve these equations simultaneously to find r and θ. Help to solve? Cant understand the question very well. I think the...

I had initially got that with my rough working out, but was confused as there are 5 terms in the square root! So the other factors would be $$(x-2)(x+3)$$ :D

Question A function $$f\left(x\right)$$ is defined by f\left(x\right)=x^{4}+4x^{3}-xx^{2}-16x-12 a) Show that there is no remainder when $$f\left(x\right)$$ is divided by $$(x+1)$$ b)Use the factor theorem to show that $$(x+2)$$ is a factor of $$f\left(x\right)$$ c) Using answers to a) and b)...

Gotten the angles correct, however got the area wrong. I used this formula, $$\sqrt{s(s-a)(s-b)(s-c)}$$ Where s = $$\frac{a+b+c}{2}$$ $$\therefore s = \frac{5.1+4.2+3.5}{2}$$=6.4 $$\sqrt{6.4(6.4-5.1)(6.4-4.2)(6.4-3.5)}$$=7.28 $$\approx7.3m$$