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Hi Melody, thank you so much for all your help!! You have really helped me out a lot! 

are you sure thats right

sqrt3/2 + sqrt3/2  =  2 (sqrt3 / 2)  = sqrt 3  = 1.732

the answer will be posted tomorrow or Tuesday fell free to use a caculator

Well,  cos(56) = .55919   so

.55919 = ab/72

or

72(.55919) = ab = 40.262

Another approach is to use similar triangles to get the height of the cylinder:

height of cylinder/(base radius of cone - radius of cylinder) = height of cone/base radius of cone

or

h/(4 - r) = 12/4

h = 3(4 - r) → 12 - 3r

V = pi*r^2*h → pi*(12r^2 - 3r^3)

or. V = 12pi*r^2 - 3pi*r^3

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Do this by equating ratios, ensuring you use consistent units.

oj/100 = 100/4000. (There are 1000 ml in a litre)

oj = 2.5 ml

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If we call as R(h), cone radius al a function of height h, then
R(h)= 4 -h/3  When cylider touchs the cone:

r = 4 - h/3 (1)

Area  A of base cylinder is:

A =¶ *r²  and h by (1) is h = (12 -3r)

Cylinder volme is:

V = A *h = ¶ *r² *(12 -3r)= ¶*(12 r^{2 -} 3r³⁾ 

In your answer ¶ is missing

Regards

Inside distance = 2 π r meters. Where r is inner radius of the track.

Outside distance = 2 π(r+5) meters

Hence extra distance = 10 π meters

Time taken for extra distance =  π seconds

Horse's speed = distance/time = 10 π/ π → 10 meters/second

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2.5

You multiply 2/3 x 3 =6/3 =2. So, h = h^2. And 4^3 = 64. So the whole thing becomes: 64h^2.

My third line above would be better written as:

 (4h^(2/3))^3 →  4^3*(h^(2/3))^3 → 64(h^(2/3))^3

Note that (a*b)^3 = (a^3)*(b^3)

With a = 4 and b = h^(2/3) we have

 (4h^(2/3))^3 → 64 (h^(2/3))^3

Also: (h^b)^c = h^(b*c)

with b = 2/3 and c = 3 we have b*c = 2, so (h^(2/3))^3 → h^2

Hence (4h^(2/3))^3 → 64h^2

Can you explain how you got that answer please?

Just noticed that Tiggsy has given the same solution.

These two solutions may be obtained by letting y = 2^sqrt(5x^2 - 8x) so the equation becomes:

8y^2 + 4 = 33y

(8y - 1)(y - 4) = 0

y = 1/8, or y = 4

y = 2^(-3) or y = 2^2

We can't have a square root being negative in the real number domain, so we must have:

sqrt(5x^2 - 8x) = 2

5x^2 - 8x  = 4

5x^2 - 8x - 4 = 0

(5x + 2)(x - 2) = 0

x = -2/5 or x = 2

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64h^2 

Choosing to read the "dots" as multiplications, there are two real solutions:

Time period, T = 1/f

2(x-4)=4x-(2x+4)

-9/7+......=1/2

bruce and sharon are the same age and both had pension plans with their place of employment . sharon started contributing $2500 per year at age 25 . bruce did not start making regular contributions until age 40 , but at that point contributed $6500 per year . sharon's plan earned and annual rate of 4.55% compounded quartely whereas bruce's plan paid an annual rate of 4.85% compounded semi-annually .

a. what was the value of each plan when each is 65 years old?

b. how much bruce and sharon contribute to their respective plans?

c. how much interest did each earn ?

d. compare their rate of returns.

ANSWERS:

This is the formula you would use to calculate this: FV=P{[1 + R]^N - 1/ R}=FV OF $1 PER PERIOD,Where R=Interest rate per period, N=number of periods, P=periodic payment, FV=Future value.

a. Sharon will have accumulated a total of: $275,957.69 when she is 65.

    Bruce will have accumulated a total of: $306,354.08 when he is 65.

b. Sharon's total contributions to her plan amounted to:$2,500 x 40 =$100,000.

    Bruce's total contributions to his plan amounted to:$6,500 x 25 =$162,500.

c. Sharon earned:$275,957.69 - $100,000 =$175,957.69 in interest.

    Bruce earned:$306,354.08 - $162,500 =$143,854.08 in interest.

d. Sharon's EFFECTIVE annual return is:4.63%.

    Bruce's EFFECTIVE annual return is:4.91%.

I choose to read this as

\(\displaystyle 2.4^{\sqrt{(5x^{2}-8x)}+1}+4 = 33.2^{\sqrt{(5x^{2}-8x})}\).

Start by making a substitution, let \(\displaystyle z = \sqrt{5x^{2}-8x}\)

and the equation becomes

\(\displaystyle 2.4^{z+1}+4=33.2^{z}\).

Next write the first 4 as 2 squared and simplify,

\(\displaystyle 8.2^{2z}-33.2^{z}+4=0\).

That's a quadratic in 2^z, so let y = 2^z and we have

\(\displaystyle 8y^{2}-33y+4=0\),

\(\displaystyle (8y-1)(y-4)=0\),

so y = 1/8 or y = 4.

Now back substitute to find z and then back substitute to find x, (and you have to solve another quadratic to do that,

i.e     \(\displaystyle 5x^{2}-8x=z^{2}\)).

You should find that x = -2/5 or x = 2.

 Tiggsy.

But note that your formula is incorrect!!

cos2A is  2cos^2(A) - 1, not 1 - 2cos^2(A)