Will85237

It is given that they are a junior, so we will only look at the junior row.

\(\frac{16}{16+2+8} \)

\(=0.615\)

Took me a while and I don't think I have the easiest method, but I've got an answer for question 2.

Let angle ABC (and all angles equal to it) be α

This allows us to make some deductions:

1)  angle AEB = 180°-α (supplementary to angle CED), and angle BAE = 180°-2α (sum of angles in isoceles triangle ABC), implies that angle ABE = 180°-(180°+α+180°-2α) = 3α-180° (sum of angles in triangle ABE), implies that angle CBD = α-(3α-180°) = 180°-2α (adjecent to angle ABE)

2) DE = 8-2 = 6 (isosceles triangle BCE)

3) if angle ABC = angle CDE and angle ACB = angle DCE, triangles ABC and CDE are similar

Using deduction 1, figure out the length of CD in terms of α using the cosine rule.

\(CD^2=BD^2+BC^2-2\times BD\times BC\times cos(CBD)\)

\(CD^2=2^2+8^2-2\times 2\times 8\times cos(180°-2α)\)

\(CD=\sqrt{68+32cos(2α)}\)

CD = EC (isoceles triangle CDE)

Using deduction 2, apply the cosine rule to triangle CDE to find a value for α (if you have a fear of long or ugly equations, look away now).

\(DE^2=CD^2+CE^2-2\times CD\times CE\times cos(DCE)\)

\(6^2=\sqrt{68+32cos(2α)}^2+\sqrt{68+32cos(2α)}^2-2\times \sqrt{68+32cos(2α)}\times \sqrt{68+32cos(2α)}\times cos(180°-2α)\)

\(0=100+64cos(2α)+200cos(2α)+64cos^2(2α)\)

Then solve the quadratic however you want (graph, formula, PlySmlt2)

\(cos(2α)=-0.625\)

\(α=64.341°\)

Use this value to find the length of CE and CD.

\(CD=\sqrt{68+32cos(2α)}\)

\(CD=\sqrt{68+32cos(2\times64.341°)}\)

\(CD=6.9282\)

Using deduction 3, find the length of AB with similar triangle rules.

\(\frac{DE}{BC}=\frac{EC}{AB}\)

\(\frac{6}{8}=\frac{6.9282}{AB}\)

\(AB=9.24\)

Feel free to ask if any of my thinking is unclear or I've made a mistake.

Apply the formula F=ma. Since mass isn't given, I will assume this car weighs 2,000 kg.

\(F=ma\)

\(F=2000\times2\)

\(F=4000N\)

Apply algebra and play with it for a bit.

\(a(x-y)=b(y-x) \)

\(ax-ay=by-bx\)

\(ax+bx=by+ay\)

\(x(a+b)=y(a+b)\)

\(x=y\frac{a+b}{a+b}\)

\(x=y\)

Therefore,

\(\frac{x}{y}=1\)

121 and 36

You can't. You need at least three known variables.

You have initial velocity and acceleration, so you will need either displacement or final velocity to find time.

\((1/2)(x)(6+x)=36\)

\((x)(6+x)=72\)

\(x^2+6x-72=0\)

\((x+12)(x-6)=0\)

\(x=-12\space or\space x=6\)

Firstly, find the angle between the ground and the ramp using tan().

\(x=arctan(\frac{opposite}{adjacent})\)

\(x=arctan(\frac{2.7}{9})\)

\(x=16.7°\)

Now use tan() again to find the height of the support.

\(tan(x)\times adjacent=opposite\)

\(tan(16.7)\times 5.5=opposite\)

\(opposite=1.65m\)

First, you need to calculate the number of moles of NaOH.

\(moles=mass/molecular\space mass\)

Na has an atomic mass of 23, O has an atomic mass of 16, and H has an atomic mass of 1. The molecular mass will be the sum of these atomic masses.

\(moles=34/(23+16+1)\)

\(moles=0.85mol\)

Each mole is 6.02*10^23 molecules, so to find the number of molecules, we multiply this number by the number of moles.

\(molecules=0.85\times6.02\times10^{23}\)

\(molecules=5.12\times10^{23}\)

There are therefore 512,000,000,000,000,000,000,000 molecules in just 34g of sodium hydroxide.

By "square 32", I'm assuming you mean "the square root of 32"?

\(\sqrt{32}\)

Find all the factors of 32.

\(\sqrt{2\times2\times2\times2\times2}\)

Group the pairs of factors together.

\(\sqrt{(2\times2)\times(2\times2)\times2}\)

Seperate the them into their own radicals.

\(\sqrt{2\times2}\times\sqrt{2\times2}\times\sqrt{2}\)

And simplify.

\(4\times\sqrt{2}\)

\(32=4\sqrt{2}\)