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These calculations are based on you working day and night, 24 hours a day, for 5 full days.

5 days x 24 hours =120 hours

120/2 = 60 hours of pay.

60 x $5.00 =$300.00 you would make in 5 days.

-sqrt(x)=64 is the same as: sqrt(x) = -64

Solve for x:
-sqrt(x) = 64

Multiply both sides by -1:
sqrt(x) = -64

Raise both sides to the power of two:
x = 4096

-sqrt(x) ⇒ -sqrt(4096) = -64:
So this solution is incorrect

The solution is incorrect:
Answer: |(no solutions)

log₁₀(log₁₀N) = 1     --->     log₁₀N = 10¹     --->     log₁₀N = 10     --->     N = 10¹⁰

loge(logeN) = 1     --->     logeN = e¹     --->     logeN = e     --->     N = e^e    

\(\frac{1}{x} + \frac{1}{y} = \frac{1}{12}\)

This is equivalent to

\(xy - 12x -12y = 0\)

Add 144 to both sides to get

\((x-12)(y-12) = 144\)

It's not hard to check all of the ways to split 144 into a product of two integers. The best we can do without allowing \(x=y\) is

\((x-12)(y-12) = 144 = 9 \times 16\)

So \(x = 28\) and \(y = 21.\)

Building on geno's answer, here is another approach.......

Whenever we have an isosceles triangle that has a base = height, any inscribed rectangle will always have the same perimeter.......and this perimeter  = twice the height of the triangle = twice the base of the triangle

Proof :

Let the base of any inscribed rectangle = 2x  and [per geno's answer], let the height  be  -2x + 8

Then......the perimeter for any x  in the domain will be =  2 [ length of rectangle base + height of rectangle ]  = 

2 [ 2x + (-2x) + 8]   =  2 [ 8]  = 16 units

Note that this is twice the base/height of the triangle......and it can be proved [ by using calculus ]  that the area of any rectangle with a fixed perimeter, P, will be maximized when each side  = P/4

So....the area of the inscribed rectangle  = [P/4]^2  =  [16/4]^2   =  4^2  = 16 units^2

Depends on how much you paid for the first sinus.....generally a discount is given for the cosinus.

....and buy a lawnmower.

p = (790 / 100) * 7

centripetal force provided by gravitational interaction:

\(\frac{mv^2}{r+h}=\frac{GMm}{(r+h)^2}\)

m = satellite mass, M = planet mass, r = planet radius, h = satellite height, G = gravitational constant, v = satellite velocity.

Rearrange:  \(GM=v^2(r+h)\)

Satellite velocity:  \(v=\frac{2*\pi(r+h)}{T}\)   T = orbital period.

Hence:  \(GM=\frac{(2\pi)^2(r+h)^3}{T^2}\)

Weight on surface:   \(W=\frac{GMm}{r^2}\rightarrow \frac{(2\pi)^2m(r+h)^3}{r^2T^2}\)

With m in kilograms, r and h in metres, T in seconds, then W is in Newtons.

I'll leave you to do the number crunching.

(4+6+8+...+2n-4)+(4+6+...+2n-6)+...+4 = (n^3 - 3n^2 - 4n + 12)/3.      n >= 4

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Here's the first one:

\(f\left( x\right)=2+x^3(7\ln(x)-9) \\ \\ f'(x)=3x^2(7\ln(x)-9)+\frac{7x^3}{x}\rightarrow x^2(21\ln(x)-20) \\ \\ f'(1)=1^2(21\ln(1)-20)\rightarrow -20 \\ \)

"Why does 0.999 equal 1."

As written it doesn't!  However, if you mean the 9's go on forever then:

Let x = 0.999999...... etc.

  10x = 9.999999..... etc.

Subtract the first from the second to get:   9x = 9  or x = 1

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If your equations are:

2y = x + 2               (1)

x - 3y = -5              (2)

Then:

Rearrange (2) to get:   x = -5 + 3y     (3)

Replace x in (1) using (3):   2y = -5 + 3y + 2  

Rearrange: -y = -3  so   y = 3

Put this into (3):    x = -5 +3*3   so  x = 4

(x, y) = (4, 3)

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If two triangles are similar, the ratio of their areas equals the ratio of the squares of their corresponding sides.

     Area₁ / Area₂  =  side₁² / side₂²

--->   270 / x  =  25² / 35²

--->   270 / x  =  625 / 1225          Cross multiply:

--->   1225·270  =  625·x

--->   330 750  =  625x                 Divide by 625:

--->     x  =  529.2

Notice that the sides are squared but the areas aren't.

Let me try part d     ---     but I'm guessing at what you mean ....

i)     f(x)  =  2 + x³(7·ln(x) - 9)          Using the product rule:

       f'(x)  =  2 + x³(7/x) + 3x²(7·ln(x) - 9)

       f'(x)  =  2 + 7x² + 3x²(7·ln(x) - 9)

       f'(1)  =  2 + 7·1² + 3·1²(7·ln(1) - 9

              --->   f'(1)  =  2 + 7 + 3(7·0 - 9)   =   2 + 7 - 27   =  -18

ii)    g(x)  =  1 + 2x²e^x + x²                        [Is this right?]

       g'(x)  =  0 + 2x²e^x + 4x·e^x + 2x

       g'(x)  =   2x²e^x + 4x·e^x + 2x

              --->   g'(1)  =  2·1²·e¹ + 4·1·e¹ + 2·1  =  2e + 4e + 2  =  6e + 2

iii)   h(x)  =  sqrt(4) + x⁴                 [Is this right?]

       h'(x)  =  0 + 4x³

       h'(x)  =  4x³

              --->   h'(1)  =  4·1³  =  4

2/sqrt(5) =2 x [sqrt(5)/5]

Why don't you try it!!. Just enter: 1/2 + 1/4 + 2/3. You should get an answer of:1.41667.....=1 5/12.

Let me try part e:     f(x)  =  x³ - 2x² - 4x + 5

i)     f'(x)  =  3x² - 4x - 4

       f''(x)  =  6x - 4

ii)     The critical points occur where the first derivate either does not exist or it equals 0.

        Since the first derivative is always defined, we need to find where it equals 0.

               f'(x)  =  3x² - 4x - 4  =  0     --->     3x² - 4x - 4  =  0     --->     (3x + 2)(x - 2)  =  0

                                                         --->      x  =  -2/3     or     x = 2           <---     the critical points

iii)     A local maximum occurs where the first derivative equals 0 and the second derivative is less than 0.

              f'(-2/3)  =  0

              f''(-2/3)  =  6(-2/3) - 4  =  -8         <---     Therefore, a local maximum occurs where  x = -2/3.

iv)     A local minimum occurs where the first derivative equals 0 and the second derivative is greater than 0.

              f'(2)  =  0

              f''(2)  =  6(2) - 4  =  8          <--     Therefore, a local minimum occurs where x = 2.

The only accurate way of comparing two similar investments, such as these two bonds, is to figure out their rate of return, which is called the "yield to maturity." There is no direct solution for the yield of a bond, but iteration and interpolation are used to find a rate that will equate the PV of the bond at maturity, which is almost always its par value, or $100.00, plus the PV of all the coupons to the purchase price of that bond.

My computer already has programmed in it the necessary software to calculate such yields almost instantaneously. All you do is enter the known variables and it will solve for the unknown, in this case, the yield to maturity:

1) A 5% coupon, paid semi-annually, 10-year bond priced @ $90.00 per hundred.

My computer calculates the yield to maturity as being: 6.36723930355% comp. semi-annually.

2) A 6% coupon, paid semi-annually, 10-year bond priced @ $97.25 per hundred.

My computer calculates the yield to maturity as being: 6.37614438663% comp. semi-annually.

As can be seen, there is virtually no difference between the two bonds, albeit the 2nd one has ever so slightly higher yield than the 1st. bond.

The solution to the system of equations:

     y  =  1.5x - 4

     y  =  -4

Replace the y-term in the first equation with -4:

     -4  =  1.5x - 4           Add 4 to both sides:

      0  =  1.5x                 Divide both sides by 1.5:

       0  =  x

The solution is:  x = 0,  y = -4