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.

$4\times\sqrt{2}$

$32=4\sqrt{2}$