Test 2 – Math 96 Flash Cards. Test 2 Functions: Domain & Range Systems of Equations Word Problems – Mixture, investment & DRT Polynomials: add, subtract,

Test 2 – Math 96

Flash Cards

Test 2

Functions: Domain & Range

Systems of Equations

Word Problems – Mixture, investment & DRT

Polynomials: add, subtract, multiply, & divide

Factor

Does the graph represents a function. State the domain and the range.

Does the graph represents a function. State the domain and the range. Vertical line test - if a vertical line will only

touch the graph at most 1 time – then the graph represents a function.

Domain: how far left , how far right Range: how low, how high

Included use [ ] Not included use ( )


Domain:

Range:

Function?


Domain:

(-4, 5]

Range:

(4, -3]

Function?

Yes

Solve this system of equations by using the Substitution method.


Isolate 1 variable in 1 of the equations;

substitute that information into the other equation.

Solve for the value of the other variable.

Go back and evaluate with the isolated variable.


Solve: x 2y 3

x y 2


Solve: x 2y 3

x y 2

2y 3 y 2

y -y

y 3 2

+3 +3

y 1

x 2( 1) 3

x 1

( 1,1)

Solve this system of equations by using the Addition/Elimination method.


If necessary, multiply one of both equations by a number so that when the two equations are added together one of the variables is eliminated.

Solve for the remaining variable; go back to one of the original equations – substitute in the value and solve for the other variable.


Solve 2x 3y 0

5x 2y 19


1) multiply by 2: 2x 3y 0

2 ) multiply by 3: 5x 2y 19 3,-2

4x 6 y 0 2( 3 ) 3y 0

15x 6 y 57 6 3y 0

19x 57

3y 6

x 3 y 2

Solve system of equations by either the Substitution or Addition method.

Solve system of equations by either the Substitution or Addition method.

To use the substitution method if a variable is isolated.

To use the addition method when both variables are on the same side of the equation.

Investment word problem, set up a system of equations & solve it.

Investment word problem, set up a system of equations & solve it. 1st equation – amounts: what was invested.

2nd equation – value: sum of percentages time amount equal the interest.

Solve by one of the above methods.

Investment word problem, set up a system of equations & solve it. Bret has $50,000 to invest, some at 12% and

the rest at 8%. If his annual interest earned is $4,740, how much did he invest at each rate?

8 50000

100 .12 .08 4740

8 8 400000

12 8 474000

4 74000

$18500 12%

$31500 8%

multiply by x y

multiply by x y

x y

x y

x

x invested at

y invested at

Investment word problem, set up a system of equations & solve it. Bret has $50,000 to invest, some at 12% and

the rest at 8%. If his annual interest earned is $4,740, how much did he invest at each rate?

8 50000

100 .12 .08 4740

8 8 400000

12 8 474000

4 74000

$18500 12%

$31500 8%

multiply by x y

multiply by x y

x y

x y

x

x invested at

y invested at

For the following mixture word problem, set up a system of equations and solve it.

For the following mixture word problem, set up a system of equations and solve it. 1st equation – amounts: ounces, pounds,

liter, etc.

2nd equation – value: $ or % multiplied to the amount added together and equal to the $ or % multiplied to the total amount.

Solve by one of the above methods.


Victor mixes a 20% acid solution with a 45% acid solution to create 100 ml of a 40% acid solution. How many ml of each solution did he use?

100 20

.20 .45 40 100

20 20 2000

20 45 4000

25 2000 80 45%

20 20%

x y multiply by

x y multiply by

x y

x y

y y ml of acid

x ml of acid


Victor mixes a 20% acid solution with a 45% acid solution to create 100 ml of a 40% acid solution. How many ml of each solution did he use?

100 20

.20 .45 40 100

20 20 2000

20 45 4000

25 2000 80 45%

20 20%

x y multiply by

x y multiply by

x y

x y

y y ml of acid

x ml of acid

For the following D = RT word problem, set up a system of equations and solve it.

For the following D = RT word problem, set up a system of equations and solve it. Label 2 equations:

Formula: (rate) (time) = distance

mph hour miles

total time ( t ) and (total – t)

total distance ( d ) and (total – d)


A sailboat leaves the island traveling at 35 mph. Five hours later a hydrofoil begins to pursue the sailboat at a speed of 60 mph. How far from the island will they meet?


A sailboat leaves the island traveling at 35 mph. Five hours later a hydrofoil begins to pursue the sailboat at a speed of 60 mph. How far from the island will they meet?

35

60 5

35 60 300

60 60 35 12

25 300 420

1 125 300 12

25 25

sailboat : d t

hydrofoil : d t

t t substitution d d

t t d

t d miles

t t

Solve this system of equations with 3 unknowns.


Use the elimination method and 2 of the original equations to make equation 4.

Use the other original equation and an already used original equation to eliminate the same variable and make equation 5.

Use equations 4 and 5 to eliminate another variable; put value back into 4 or 5, put both values back into and original equation.


1) x + y + z = 8 2) x + 2y – z = 3 3) 2x – y + z = 3


1) x + y + z = 8 2) x + 2y – z = 3 3) 2x – y + z = 3

1) x + y + z = 8 2) x + 2y – z = 3 4) 2x + 3y = 11

2) x + 2y – z = 3 3) 2x – y + z = 3

5) 3x + y = 6

4) 2x + 3y = 115) -9x – 3y = -18 -7x = -7 -7 -

7 x = 1

2x + 3y = 112(1) + 3y = 11 -2 -2 3y = 9 3 3 y = 3

Solve for z x + y + z = 8 (1) + (3) + z = 8

4 + z = 8-4 -4

z = 4

Add or Subtract the polynomials as indicated. Simplify your answer.

Add or Subtract the polynomials as indicated. Simplify your answer. Combine like terms −

Beware of subtraction outside of the parentheses, change all the signs inside the parentheses.


2 2x 3x 4 2x x 7


2 2x 3x 4 2x x 7

2 2x 3x 4 2x x 7

2x 4x 11

Multiply the following polynomials as indicated. Simplify your answer.

Multiply the following polynomials as indicated. Simplify your answer. If there is an “outside” exponent, write the

base as many time as indicated by the exponent.

Use the distributive property.

Simplify your answer by combining like terms.


2

2

2

2x 5

2x 5 2x 5

4x 10x 10x 25

4x 20x 25


2

2

2

2x 5

2x 5 2x 5

4x 10x 10x 25

4x 20x 25

Use long division, where R is less than the degree of D.

Use long division, where R is less than the degree of D. Divide the leading term of the denominator

into the leading term of the numerator.

Multiply back to all terms that were in the denominator.

Subtract, better yet, add the opposite.

Repeat above steps till degree “inside” is less than the degree “outside”.


22

2

2

3x 26 x 5x 1

2x 3 6 x 5x 12x 3

6 x 9x

4x 1

4x 6

5

6 x 5x 1 53x 2

2x 3 2x 3


22

2

2

3x 26 x 5x 1

2x 3 6 x 5x 12x 3

6 x 9x

4x 1

4x 6

5

6 x 5x 1 53x 2

2x 3 2x 3

Factor completely – 2 Terms

Factor completely – 2 Terms Always look for a common factor

immediately take it out to the front of the expression all common factors, show what’s left inside ONE set of parenthesis

Identify the number of terms. If there was a common factor, then look at the number of terms left inside the parenthesis.

If there are only two terms, see if it is the sum or difference of perfect cubes or squares

a2 – b2 = (a + b)(a – b) a3 – b3 = (a – b)(a2 + ab + b2) OR a3 + b3 = (a + b)(a2 – ab + b2)

To factor the sum and difference of cubes, write what you see without exponent 3, “square” “smile” “square”, the second sign is different, the last sign is always plus.

Factor completely: 40 – 5y3

Factor completely: 40 – 5y3

Common: 5(8 – y3)

See exponent 3: 5(23 – y3)

5(2 – y)(4 + 2y + y2)

Factor completely – 3 Terms

Factor completely – 3 Terms Always look for a common factor

immediately take it out to the front of the expression all common factors

show what’s left inside ONE set of parenthesis Identify the number of terms. If there are three terms, and the leading coefficient is positive:

find all the factors of the first term, find all the factors of the last term Within 2 sets of parentheses,

place the factors from the first term in the front of the parentheses place the factors from the last term in the back of the parentheses

NEVER put common factors together in one parenthesis. check the last sign,

if the sign is plus: use the SAME signs, the sign of the 2nd term if the sign is minus: use different signs, one plus and one minus

“smile” to make sure you get the middle term multiply the inner most terms together then multiply the outer

most terms together, and add the two products together.

Factor completely: 2x2 – 5x – 7

Factor completely: 2x2 – 5x – 7

Factors of the first term: 1x & 2x

Factors of the last term: -1 & 7 or 1 & -7

(2x – 7)(x + 1)

Factor completely : 4 Terms

Factor completely : 4 Terms

Always look for a common factor immediately take it out to the front of the expression all

common factors, show what’s left inside ONE set of parenthesis

Identify the number of terms. If there was a common factor, then look at the number of terms left inside the parenthesis.

Do NOT remove parentheses already given in the expression.

If there are four terms: try to factor the first three terms if perfect square trinomial

write it as (ax + b)2 – y2 factor as the difference of perfect squares.

try to factor the out a negative from the last three terms write it as x2 – (ay + b)2 factor as the difference of perfect squares

Factor completely: x2 – y2 + 10y – 25

Factor completely: x2 – y2 + 10y – 25

Factor a negative out of the last 3 terms

x2 – (y2 – 10y + 25)

x2 – (y – 5)(y – 5)

x2 – (y – 5)2

(x + (y – 5))(x – (y – 5))

(x + y – 5)(x – y + 5)

Documents

Test 2 – Math 96 Flash Cards. Test 2 Functions: Domain & Range Systems of Equations Word Problems – Mixture, investment & DRT Polynomials: add, subtract,