i really need your help guys answering this problems so i can review it

1. A pasture is twice as long as it is wide. Its area is 115,200 ft^2. How wide is the pasture?

Let the width be "w. The pasture is "twice as long as it is wide" so its length is "2w. Since the area of a rectangle is "length times width", \(\displaystyle (2w)w= 2w^2= 115200\). Solve that equation.

2.Find the length x in the figure, if the shaded area is 144cm^2.

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If you continue that "middle" line downward, you divide the figure into two rectangles, one "10 by x" and the other "6 by x". They have area 10x and 6x respectively. The total area is 10x+ 6x=144.

3. A health clinic uses a solution of bleach to sterilize petri dishes in which cultures are grown. The sterilization tank contains 100 gal of a solution of 2% ordinary household bleach mixed with pure distilled water. New research indicates that the concentration of bleach should be 5% for complete sterilization. How much of the solution should be drained and replaced with bleach to increase the bleach content to the recommended level?

If you drain of "x gallons" of 2% bleach you will have left 100- x gallons of mixture which means you will have .02(100- x) gallons of pure bleach. If you now add x gallons of pure bleach, you will have .02(100- x)+ x= 2+ .98x gallons of pure bleach. For that to be a 5% solution, that must equal 5 gallons:

solve 2+ .98x= 5.

4. A merchant blends tea that sells for $3.00 a pound with tea that sells for $2.75 a pound to produce 80lb of a mixture that sells for $2.90 a pound. How many pounds of each type of tea does the merchant use in blend?

Let "x" be the number of pounds of the first type of tea in the mixture. Then there will be 80- x pounds of the other kind of tea. The "x" pounds of the first tea would sell for 3x dollars and the 80-x pounds of tea would sell for 2.75(80- x) dollars. At 2.90 a pound, those 80 pounds would sell for 80(2.90 so you want 2x+ 2.75(80- x)= 80(2.90). Solve that for x.

5. A wood cutter determines the height of a tall tree by first measuring a smaller one 125ft away, then moving so that his eye are in the line of sight along the tops of the tree, and measuring how far he is standing from the small tree. Supposed the small tree is 20ft tall, the man is 25ft from the smaller tree, and his eye level is 5 ft above the ground, how tall is the taller tree?

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By "moving so that his eye

**is** in the line of sight along the tops of the trees", the wood cutter is setting up "similar triangles"- all angles are the same so corresponding sides are in equal ratios. The base of those triangles is at his eye so the top of the small tree is 20- 5= 15 feet above his eye. The small tree is 25 feet from the wood cutter so "height (above his eye) over distance" is \(\displaystyle \frac{20}{25}\). The taller tree is 125+ 25= 150 feet from the wood cutter at this time, so, letting "x" be "height above his eye" for the taller tree, "height (above his eye) over distance" is \(\displaystyle \frac{x}{150}\). Since those ratios are the same,

\(\displaystyle \frac{x}{150}= \frac{20}{25}\). Don't forget to add the 5 feet for the height to his eye to get the height of the tree above the ground.

6.A plank 30ft long rests on top of a flat-roofed building, with 5 ft of the plank projecting over the edge, as shown in the figure. A worker weighing 240 lb sits on one end of the plank. What is the largest weight that can be hung on the projecting end of the plank if it is to remain in balance?

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The physical principal here is that different "torques", distance times force, about the pivot point, where the plank crosses off the roof, must be equal. Let the weight be "w". Since it is 5 feet off the roof, its torque about that point is 5w. The worker is at the other end of the plank, 20- 5= 15 feet from the pivot point. He weighs 240 pounds so his torque about the pivot point is 15(240). To be equal, you must have 5w= 15(240). Solve that for w.

7. A woman driving a car 14ft long is passing a truck 30ft long. The truck is traveling at 50mi/h. How fast must the woman drive her car so that she can pass the truck completely in 6 s, from the position shown in figure (a) to the position shown in figure (b)? use feet and seconds instead of miles and hours.

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In order to completely pass the truck, the rear of her car must be past the truck so the front of her car must go 30+ 14= 44 feet. Since we want to "use feet and seconds instead of miles and hours, convert 50 mph to feet per second: there are 5280 feet in a mile to 50 miles is 50(5280). There are 60 minutes in an hour and 60 seconds in a minute so one hour is 60(60) seconds. 50 "miles per hour" is 50(5280) feet per 60(60) seconds or \(\displaystyle \frac{50(5280)}{60(60)} feet per second. If you let "v" be her speed in feet per second,, the her speed "relative to the truck" is \(\displaystyle v- \frac{50(5280)}{60(60)}\) feet per second. At that relative speed, in 6 seconds she will have gone \(\displaystyle (v- \frac{50(5280)}{60(60)})6\) feet and that must be 44: Solve \(\displaystyle (v- \frac{50(5280)}{60(60)})6= 44\) for v.

8. A running track has the shape shown in the figure, with straight sides and semicircular ends. If the length of the track is 440 yd and the two straight parts are each 110 yd long, what is the radius of the semicircular parts?(to the nearest yard)?

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If the entire track is 440 yds and the two straight parts are each 110 yards long that leaves 440- 110- 110= 220 yards for the two semi-circular parts. A circle with radius r has circumference \(\displaystyle 2\pi r\) and the two semi-circular ends form one circle with circumference 220. Solve \(\displaystyle 2\pi r= 220\).

thanks for helping

May I ask why you appear to have made no attempt at these problems yourself? Whoever gave you these problems clearly expects that you already know these concepts. If that is not the case, I urge you to let your teacher know that!\)