Alan

 $$\frac{x}{4.8}-30=-10$$

Add 30 to both sides

$$\frac{x}{4.8}= 20$$

Multiply both sides by 4.8

$$x = 4.8*20$$

$$x=96$$

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3X + 49 = 2X + 11

Subtract 2X from both sides

X + 49 = 11

Subtract 49 from both sides

X = -38

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Let the speed of the bike going south be s km/hr.  Then the speed of the bike going north is (s+4) km/hr

The speed of the bike going north is (2s + 4) km/hr relative to the one going south.  Therefore In 1.5 hours they will be a distance (2s + 4)81.5 km apart.  This, we are told, ios 39 km.

(2s + 4)*1.5 = 39

3s + 6 = 39

3s = 33

s = 11

So the south going bike is travelling at 11 km/hr and the north going bike is travelling at 15 km/hr

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Chris beat me to it!

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The 120 feet of fencing will form 3 sides of the outdoor room.  Thw two sides touching the museum will have width W; the third side, joining these will have length L (which is what you are asked to find).

So we must have:  2W + L = 120  or  W = 60 - L/2

The enclosed area is given by A = W*L  or  A = (60 - L/2)*L  or 60*L - L²/2

To find the maximum area we differentiate A with respect to L and set the result to zero:

dA/dL = 60 - L

For dA/dL to be zero we must have L = 60

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Tackle as follows:

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001.

Use v = u + at  (assuming constant acceleration applied by the bowstring).

127 = 0 + 139t

t = 127/139 s

$${\mathtt{t}} = {\frac{{\mathtt{127}}}{{\mathtt{139}}}} \Rightarrow {\mathtt{t}} = {\mathtt{0.913\: \!669\: \!064\: \!748\: \!201\: \!4}}$$

t ≈ 0.9137 s

002.

Since it doesn't accelerate after it leaves the bow and there is nothing to retard it in space, it continues with the same speed: v = 127 m/s.

003.

distance = speed*time

distance = 127*3.5 m

$${\mathtt{distance}} = {\mathtt{127}}{\mathtt{\,\times\,}}{\mathtt{3.5}} \Rightarrow {\mathtt{distance}} = {\mathtt{444.5}}$$

distance = 444.5m

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First question.  These graphs might help:

In general, insert the two values into the function.  If the results are of opposite sign there is a zero between them.

Second question

Use  tangent:  tan(angle) = opposite/adjacent

Here:  tan(41°) = height/1210

so:  height = 1210*tan(41°)

$${\mathtt{height}} = {\mathtt{1\,210}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{41}}^\circ\right)} \Rightarrow {\mathtt{height}} = {\mathtt{1\,051.836\: \!952\: \!757\: \!36}}$$

height = 1052 ft to the nearest foot.

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mean radius R = 7926*1.6093/2 = 6378 km to nearest km.

Angle C = 90°+30° = 120°

legnth AB = 6378 + 2000 = 8378 km

sin(A)/6378 = sin(C)/8378       so    sin(A) = sin(C)*6378/8378 = sin(120°)*6378/8378 = 0.69529

Angle A = sin-1(0.69529)  ≈ 41°

Angle B = 180° - C - A = 180° - 120° - 41°  =  19°

B/360° = t/2  where t is time in hours

t = 2*19/360 hrs = 0.19hrs

t = 6.333... mins = 6 mins 20 secs

12 noon - 6.33 mins = 11.53am + 40 secs.

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Here's a)

You can use the same method on b).

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