1.Find the discounted price of a tent with a price of $89 and a discount of 15%. 2.Find the final...
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Transcript of 1.Find the discounted price of a tent with a price of $89 and a discount of 15%. 2.Find the final...
1. Find the discounted price of a tent with a price of $89 and a discount of 15%.
2. Find the final price of a pair of hiking boots with a price of $78, a discount of 10%, and a tax of 6%.
3. On September 1, a stock sold for $46 per share and on October 1 it sold for $48.30 per share. What was the percent of change in the price of the stock?
$75.65
$74.41
5% increase
Math 8H
Algebra 1 Glencoe McGraw-Hill JoAnn Evans
Problem Solving Day 3Rectangle/Frame Problems
A rectangular swimming pool is 4 meters longer than it is wide. The pool is surrounded
by a cement sidewalk that is 1 meter wide. The area of the sidewalk is 32 m2. Find the
dimensions of the pool.
Let Statements:Let x = width of the pool Let x + 4 = length of the poolLet x + 6 = total lengthLet x + 2 = total width
Verbal Sentence:Area of pool + area of sidewalk = total area
x (x + 4) + 32 = (x + 6) (x + 2)
x + 4
x + 4x x
1 + (x + 4) + 1
1 +
x +
1
Equation:
x (x + 4) + 32 = (x + 6) (x + 2)
4x + 32 = 8x + 12 -4x -4x 32 = 4x + 12 -12 -12 20 = 4x 5 = xSolution:
The pool is 5 meters wide and 9 meters long.
+ 4x
+ 32 = x2 + 2x +6x +12
x2
A painting is 10 cm longer than it is wide. It is mounted in a frame that is 1.5 cm wide. The area of the frame is 339 cm2. Find the
dimensions of the painting.
Let Statements:Let x = width of the paintingLet x + 10 = length of the paintingLet x + 3 = width of framed paintingLet x + 13 = length of framed painting
Verbal Sentence:Area of painting + area of frame = total area
x ( x + 10) + 339 = (x + 3) (x + 13)
x
x +
10
1.5 + x + 1.5
1.5
+ (x
+10) +
1.5
Equation:
x ( x + 10) + 339 = (x + 3) (x + 13)
x2 + 10x + 339 = x2 + 13x + 3x + 39 10x + 339 = 16x + 39 -10x -10x 339 = 6x + 39 -39 -39 300 = 6x 50 = x Solution:
The painting is 50 cm wide and 60 cm long.
A rectangle is 4 meters longer than it is wide. If the length and width are both
increased by 5 meters, the area is increased by 115 m2. Find the original dimensions.
Let Statements:Let x = original width of rectangleLet x + 4 = original length of rectangleLet x + 5 = new rectangle widthLet x + 9 = new rectangle length
Verbal Sentence:original area + 115 = new larger rectangle area
x + 4
x
(x + 4) + 5
x + 5
x (x + 4) + 115 = (x + 5) (x + 9)
Equation: x (x + 4) + 115 = (x + 5) (x + 9)
x2 + 4x + 115 = x2 + 9x + 5x + 45
4x + 115 = 14x + 45 -4x -4x 115 = 10x + 45 -45 -45 70 = 10x 7 = x Solution:
The original dimensions were 7 m wide and 11 m long.
A Mini Cooper and a truck heading for San Pedro on the same freeway left the Redwood Middle School
parking lot at the same time. The Mini Cooper, which drove 20 mi/h faster than the truck, arrived in San Pedro after two hours. The truck arrived in San Pedro one hour later than the car. Find the rate of
the car.
Let Statements:Let x = rate of Mini CooperLet x - 20 = rate of truck
Verbal Sentence: distance Mini Cooper drives = distance truck
drives
RMS San Pedrodistance = distance
Equation:
Solution:
The car’s rate was 60 mi/h.
Mini’s distance = Truck’s distance
x(2) = (x – 20)3
rt = rt
2x = 3x – 60-3x -3x-x = -60x = 60