course Mth 151
April 25 around 10:00
assignment 28-QAq001:
we would first subtract 7 from both sides which would equal 2x=14. then we divide both sides by 2 which would equal x=7.
*3
q002:
we would first perform the distributive property on the left side to get 2x+14=30. Then we would subtract 14 from both sides to get 2x=16. Then we would divide both sides by 2 to get x=8.
*3
q003:
If x stands for the number, then double the number is 2x, and double the number plus 5 is therefore 2x+5
*2
q004:
As seen in the preceding question, 'double a number plus five' can be expressed as 2x+5, so the statement 'double a number plus five is 19' can be written as the equation 2x+5=19
*2
q005:
The perimeter of a figure is the distance around it. A square consists of 4 equal sides. If each side has length x, then the distance around it is 4*x, or just 4x
*2
q006:
The area of a square is the product of the lengths of its sides. If x stands for the length of a side, then the area of the square is x*x or x^2
*2
q007:
The perimeter is the sum of the lengths of the sides of the rectangle. Since we are given the relationship between the lengths of the sides, we first express this relationship in symbols. If we let x stand for the width of the rectangle, then the length of the rectangle is 2x.
Since both the length and the width occur twice as we move around the perimeter of the rectangle, it follows that the perimeter is 2*length+2*width=2*2x+2*x. This expression simplifies to 4x+2x, which further simplifies to 6x
*2
q008:
The area of a rectangle is the product of its length then width. Since we are given the relationship between length in width, we begin by expressing the length and width of the rectangle in terms of a symbol. If we let x stand for the width of the rectangle, then 2 x stands for its length.
Now since the area is the product of length and width, we see that the area must be x*2x, or 2x^2
*2
q009:
The area of a rectangle whose length is double its width was seen in the preceding problem to be 2 x^2, where x is the width. If the area is 72, we see that 2 x^2 = 72.
We solve this equation as follows. Starting with 2 x^2 = 72 we divide both sides by 2 to get x^2 = 36. We then take the square root of both sides of the equation to obtain x=`sqrt(36), or
x = 6.
Note that the equation x^2 = 36 actually has two solutions, x = 6 and x = -6. However the length of a side cannot be negative so we chose the positive solution
*2
q010:
If x represents the amount invested at five percent, then the remaining amount is $15,000-x, which is invested at eight percent.
The annual interest on amount x invested at five percent is .05x, while the annual interest on amount ($15,000-x) invested at eight percent is .08($15,000-x). Thus the total interest on the $15,000 is total interest=.05x+.08($15,000-x)
*2
q011:
If the total interest is $1000, then knowing that the expression for the total interest is .05x+.08($15,000-x) we obtain the equation
.05x+.08($15,000-x)=$1000.
To solve the equation .05x+.08($15,000-x)=$1000 we first apply the distributive law on the left-hand side to obtain
.05x+$1200-.08x=$1000. Simplifying the left-hand side we obtain
-.03x+$1200=$1000. Subtracting $1200 from both sides we get
-.03x=-$200. Dividing both sides by -.03 we obtain{{}x=-$200/-.03=$6667, rounded to the nearest dollar.
This means that of the $15,000, the amount invested at five percent is $6667
*2
q012:
A good rule of thumb is that whenever an equation involves denominators, we multiply both sides by a quantity that eliminates the denominators. In this case if we multiply both sides by 3 we will eliminate the denominator of the term x/3.
Multiplying both sides of x/3+9=27 by 3 we obtain
3(x/3+9)=3*27. Applying the distributive law to the left-hand side and multiplying the numbers on the right-hand side this becomes
3*(x/3)+3*9=81, which simplifies to give us
x+27=81. We now subtract 27 from both sides to obtain the solution
x=54
*2
q013:
This is a quadratic equation, solve using the quadratic formula. Recall that the quadratic formula tells us that the equation ax^2+bx+c=0 has solutions x=[-b+-`sqrt(b^2-4ac)]/(2a), where the + - indicates '+ or -', meaning that both + and - symbols in that position will give us correct solutions.
In the present case we see that a = 2, b = 9, and c = -68, so that the solutions are x=[-9+-`sqrt(9^2-4*2*(-68))]/(2*2)=[-9+-`sqrt( 625)]/4=[-9+-25]/4.
Choosing the + we have x=[-9+25]/4=164=4.
Choosing the - we have [-9-25]/4=-34/4=-8.5
*2
q014:
A common denominator for this equation would be 10x. Thus multiplying both sides by 10x would eliminate all denominators. We obtain 10x(3/x+x/2)=10x(31/10). Applying the distributive law to the left-hand side we obtain 10x(3/x)+10x(x/2)=10x(31/10). Simplifying we have 30+5x^2=31x. This might appear difficult to solve, but if we subtract the 31x from both sides and rearrange the order of the left-hand side we have
5x^2-31x+30=0, which we now recognize as a quadratic equation with a=5, b=-31 and c=30. Thus are solutions are
x=[-b+-'sqrt(b^2-4ac)]/(2a)=
[-(-31)+-`sqrt((-31)^2-4*5*30]/(2*5)=
[31+-`sqrt(361)]/10=
[31+-19]/10.
Choosing the + solution we have x=[31+19]/10=50/10=5.
Choosing the - solution we have x=[31-19]/10=1.2
*2
q015:
In order to express the ratio both quantities must be given in the same units. Since 2ft. is the same as 24inches, the ratio is 24inches to 72inches, or 24in/72in=1/3.
We could alternatively have expressed the 72inches as 6ft. to get the ratio of 2ft./6ft.=1/3
*2
q016:
There are 4 quarts in a gallon and 2 pints in a quart, so there are eight pints in a gallon. In five gallons we therefore have 40 pints. The desired ratio is therefore 40 pints/72 pints=5/8
*2
q018:
A square three meters on a side has area (3 meters)^2=9m^2, while a square 2 meters on a side has area (2 meters)^2=4m^2, so the ratio of areas is 9m^2/(4m^2)=9/4
*2
q019:
A square 4 cm on a side has area (4cm)*(4cm)=16cm^2.
A square 2 meters on a side has area (200cm)*(200cm)=40,000cm^2.
The ratio of the areas is therefore 16 cm^2/(40,000cm^2)=1/2500
*2
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