Alan

Let one boat be distance d1 from the observer, and the other be at a distance d2.

 We have two right-angled triangles for which we know the angle and the opposite side (the height), so we can use the tangent function to find the adjacent sides (the distances from the observer).

tan(31°) = 95/d1  or d1 = 95/tan(31°)

similarly d2 = 95/tan(42°)

so distance between boats is d1 - d2

$${\frac{{\mathtt{95}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{31}}^\circ\right)}}}{\mathtt{\,-\,}}{\frac{{\mathtt{95}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{42}}^\circ\right)}}} = {\mathtt{52.598\: \!361\: \!914\: \!428\: \!964\: \!3}}$$

The boats are ≈ 52.6 feet apart.

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I'll do a couple for you:

A.  x + 2y = 17

     x - y    = 2

Subtract the second equation from the first, term by term

x - x + 2y - (-y) = 17 - 2

or 3y = 15

Divide both sides by 3

y = 5

Put this back into the first equation

x +2*5 = 17

x + 10 = 17

x = 17 - 10

x = 7

B. 4x + 5y = 11

    2x + 6y = 16

Double all the terms in the second equation to get the number of x's the same as in the first equation

   4x + 12y = 32

Subtract the first equation from the third equation term by term

   4x - 4x + 12y - 5y = 32 - 11

or 7y = 21

Divide by 7

y = 3

Put this back into the first equation

4x + 5*3 = 11

4x + 15 = 11

4x = -4

x = -1

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1/4 of a mile in 25 seconds is 1 mile in 100 seconds.  There are 3600 seconds in one hour so there are 36 lots of 100 seconds in one hour.  This means the guy can run 36 mph (must be the six-million dollar man!).

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$$(3x^2-4x+7)(-2x^2+5x-9)=-6x^4+23x^3-61x^2+71x-63$$

Think of this as:

$$3x^2(-2x^2+5x-9)-4x(-2x^2+5x-9)+7(-2x^2+5x-9)$$

Then multiply the 3x² by all the terms inside the brackets; do the same for the -4x and the +7; then collect like terms together.

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There are 180° in pi radians, so 1° is pi/180 radians, hence 30° is 30*pi/180 radians = pi/6 radians

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You need to know about analytic continuation to understand this.

Use v² = u² + 2*a*s  where v = final speed (0 m/s), u = initial speed (120*10³/3600 m/s) a = acceleration ( -6.5 m/s²) and s = distance (in metres).

Rearrange as: s = (v² - u²)/(2a)

$${\mathtt{s}} = {\frac{\left({\mathtt{0}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{120\,000}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{6.5}}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{85.470\: \!085\: \!470\: \!085\: \!470\: \!1}}$$

or s ≈ 85.5 m

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I agree that Chris's analysis is correct (apart from the bit that says "divide both sides by 32"  - this should say "divide both sides by 16"!!)

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mass of neutron/mass of electron = 2*10^-24/(9*10^-28)

$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{24}}\right)}}{\left({\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{28}}\right)}\right)}} = {\mathtt{2\,222.222\: \!222\: \!222\: \!222\: \!222\: \!2}}$$

so d is closest.

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The general equation for a parabola can be written as: (x - h)² = 4a(y - k)  where (h, k) are the coordinates of the vertex.

Since we are told that y = 20 at the vertex, then we know immediately that k = 20.

The y-intercept occurs where x = 0, so we have:  (0 - h)² = 4a(15 - 20)    ...(1)

A zero of the equation means the value of x where y = 0, so we also have:  (3 - h)² = 4a(0 - 20)    ...(2)

Can you use equations (1) and (2) to find two sets of values for h and a?

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