MAC 1140

Final Exam Review

Information about the final exam

• The final exam is comprehensive.

• Students who do not take the final exam will receive a zero for the final exam unless prior arrangements have been made for a rare incomplete.

• If you miss the final and do not email me by the day of the exam, I will assume you wish to take a zero on your final exam.

Exam Topics (1 – 10) Objective Test Section Suggested Text Problems

1) Find a polynomial with given zeros and y – intercept. 1:2 4.1 Page 195: 105

2) Find the complex zeros (all zeros) of a polynomial function. 1:9 4.3 Page 245: 27, 29

3) Find the vertical, horizontal, and oblique asymptotes of a rational function.

1:11 4.4 Page 245: 31 (asymptotes)

4) Solve a polynomial inequality. 1:14 4.6 Page 247: 41

5) Solve a rational inequality. 1:15 4.6 Page 247: 43

6) Find compositions of functions (and domains). 2:2 5.1 p. 348 5, 7

7) Find the inverse of a function (and domains and ranges) 2:3 5.2 p. 348: 11, 13

8) Solve a logarithmic equation. 2:8 5.6 p. 349: 37, 40, 43

9) Solve an exponential equation. 2:9 5.6 p. 349: 35, 38

10) Find the time required to reach a monetary goal. 2:11 5.7 p. 350: 50 (final part) Answer: 17.5 years

Exam Topics (11 – 20) 11) Find the vertex, focus, and directrix, and orientation of opening of a parabola.

3:1 10.2 p. 698: 1, 9

12) Write the equation of a hyperbola given the foci and the vertices

3:3 10.4 p. 699: 12

13) Solve a determinant equation. 3:6 11.3 p. 741: 51 – 56

14) Solve a curve fitting application. 3:9 11.2 – 11.4

p. 795: 55

15) Solve a system of nonlinear equations by any method. 3:10 11.6 p. 774: 71 – 82

16) Find the sum of a sequence. 4:2 12.1 p. 838: 13 – 15

17) Find a general formula for a given arithmetic sequence. 4:4 12.2 p. 838: 21

18) Work with applications of arithmetic sequences. 4:5 12.2 p. 839: 34 (a)8,(b)1100

19) Determine whether a geometric series converges or diverges (and determine the sum if it converges).

4:7 12.3 p. 838: 23, 25

20) Use the Binomial theorem to find a particular coefficient or term in a binomial expansion.

4:10 12.5 p. 836: 29 – 31

1) Find a polynomial with given zeros and y – intercept.

Multiplicity

Let m be a positive integer. If (𝑥 − 𝑟)𝑚 is a factor of a polynomial f and (𝑥 − 𝑟)𝑚+1

is not a factor of f, then r is called a zero of f of multiplicity m.

Page 195: 105

Zeros −5 (multiplicity 2); 2 (multiplicity 1);

4 (multiplicity 1); degree 4;

contains the point (3, 128)

𝑓 𝑥 = 𝑎 𝑥 − −52𝑥 − 2 1 𝑥 − 4 1

= 𝑎 𝑥 + 5 2(𝑥 − 2)(𝑥 − 4) 𝑓 3 = 𝑎 3 + 5 2 3 − 2 3 − 4 = 128

−64𝑎 = 128 𝑎 = −2

𝑓 𝑥 = −2 𝑥 + 5 2(𝑥 − 2)(𝑥 − 4)

2) Find the complex zeros (all zeros) of a polynomial function.

Finding the Zeros of a Polynomial Function • Use the degree to determine the maximum number of zeros. • Use the Rational Zeros Theorem to identify possible rational

zeros. • Graph the function to find the best choice of possible rational

zeros. Note how many real zeros are possible. • Use the graphing calculator and find the value of the function at

a possible rational zero. If the value is zero, you have found an actual rational zero.

• Find the new depressed function using synthetic division. Find another rational zero and continue this process until you have found all rational zeros or the depressed equation is a quadratic equation that can be solved.

Page 245: 29 𝑓 𝑥 = 𝑥4 − 4𝑥3 + 9𝑥2 − 20𝑥 + 20 𝑝: ±1,±2,±4,±5,±20

𝑞: ±1

𝑝

𝑞: ±1

1, ±2

1, ±4

1,±5

1,±20

1

±1,±2,±4,±5,±20 Graphing yields a zero at 𝑥 = 2. Based on the fact that there are no other real zeros (and the degree is four), it looks like 𝑥 = 2 could be a double root. Synthetic division by 𝑥 − 2 yields 𝑥3 − 2𝑥2 + 5𝑥 − 10. Dividing by 𝑥 − 2 yields 𝑥2 + 5 Solve 𝑥2 + 5 = 0

𝑥2 = −5

𝑥 = ± 5𝑖

𝑍𝑒𝑟𝑜𝑠: 2 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑖𝑡𝑦 2 ,± 5𝑖

3) Find the vertical, horizontal, and oblique asymptotes of a rational function.

Determining Vertical Asymptotes of

𝒇 𝒙 =𝒑(𝒙)

𝒒(𝒙)

1) Simplify the function completely.

2) Set the denominator equal to zero and solve the resulting equation. The line x = a is

a vertical asymptote of the function if a is a solution to that equation.

Determining Horizontal Asymptotes of

𝒇 𝒙 =𝒑(𝒙)

𝒒(𝒙)

• If the degree of the numerator is less than the degree of the denominator, the x-

axis (y = 0) is the horizontal asymptote. • If the degree of the numerator is equal to the degree of the denominator, the

line formed by the ratio of the leading coefficients is the horizontal asymptote. • If the degree of the numerator is greater than the degree of the denominator,

there is no horizontal asymptote (but if the degree is exactly one more, there is an oblique asymptote). – Find an oblique asymptote by dividing the denominator into the numerator. The result

should be linear.

Page 245: 31 (asymptotes)

𝑅 𝑥 =𝑥 + 2

𝑥2 − 9

=𝑥 + 2

𝑥 + 3 (𝑥 − 3)

𝑉. 𝐴. : 𝑥 = −3, 𝑥 = 3 𝐻. 𝐴. : 𝑦 = 0

4) Solve a polynomial inequality.

Steps for Solving Polynomial and Rational Inequalities Algebraically Step 1: Write the inequality so that a polynomial or rational expression f is on the left side and zero is on the right side in one of the following forms: 𝑓 𝑥 > 0, 𝑓 𝑥 ≥ 0 𝑛𝑜𝑡 𝑠𝑡𝑟𝑖𝑐𝑡 , 𝑓 𝑥 < 0, 𝑓 𝑥 ≤ 0(𝑛𝑜𝑡 𝑠𝑡𝑟𝑖𝑐𝑡) For rational expressions, be sure that the left side is written as a single quotient. Step 2: Determine the numbers at which the expression 𝑓 on the left side equals zero and, if the expression is rational, the numbers at which the expression 𝑓 on the left side is undefined. Step 3: Use the numbers found in Step 2 to separate the real number line into intervals. Step 4: Select a test number c in each interval and evaluate f at the test number. (a) If the value of 𝑓(𝑐) is positive, then 𝑓 𝑥 > 0 for all numbers 𝑥 in the interval. (b) If the value of 𝑓(𝑐) is negative, then 𝑓 𝑥 < 0for all numbers 𝑥 in the interval. If the inequality is not strict , include the solutions of 𝑓 𝑥 = 0 in the solution set. Be careful, however. Test numbers at which 𝑓 is not defined are not solutions to 𝑓 𝑥 = 0 and should not be included in the solution set.

Page 247: 41

𝑥3 + 𝑥2 < 4𝑥 + 4 𝑥3 + 𝑥2 − 4𝑥 − 4 < 0 𝑥2 𝑥 + 1 − 4 𝑥 + 1 < 0 𝑥 + 1 𝑥2 − 4 < 0 𝑥 + 1 𝑥 + 2 𝑥 − 2 < 0 𝑍𝑒𝑟𝑜𝑠: −2,−1, 2

−∞,−2 : 𝑓 −3 = −15 − −2,−1 : 𝑓 −1.5 = 0.875 + −1, 2 : 𝑓 0 = −4 − 2,∞ : 𝑓 3 = 20 +

−∞,−2 𝑈(−1,2)

5) Solve a rational inequality.

Steps for Solving Polynomial and Rational Inequalities Algebraically Step 1: Write the inequality so that a polynomial or rational expression f is on the left side and zero is on the right side in one of the following forms: 𝑓 𝑥 > 0, 𝑓 𝑥 ≥ 0 𝑛𝑜𝑡 𝑠𝑡𝑟𝑖𝑐𝑡 , 𝑓 𝑥 < 0, 𝑓 𝑥 ≤ 0(𝑛𝑜𝑡 𝑠𝑡𝑟𝑖𝑐𝑡) For rational expressions, be sure that the left side is written as a single quotient. Step 2: Determine the numbers at which the expression 𝑓 on the left side equals zero and, if the expression is rational, the numbers at which the expression 𝑓 on the left side is undefined. Step 3: Use the numbers found in Step 2 to separate the real number line into intervals. Step 4: Select a test number c in each interval and evaluate f at the test number. (a) If the value of 𝑓(𝑐) is positive, then 𝑓 𝑥 > 0 for all numbers 𝑥 in the interval. (b) If the value of 𝑓(𝑐) is negative, then 𝑓 𝑥 < 0for all numbers 𝑥 in the interval. If the inequality is not strict , include the solutions of 𝑓 𝑥 = 0 in the solution set. Be careful, however. Test numbers at which 𝑓 is not defined are not solutions to 𝑓 𝑥 = 0 and should not be included in the solution set.

Page 247: 43 2𝑥 − 6

1 − 𝑥< 2

2𝑥 − 6

1 − 𝑥− 2 < 0

2𝑥 − 6

1 − 𝑥−2 1 − 𝑥

1 − 𝑥< 0

2𝑥 − 6 − 2 + 2𝑥

1 − 𝑥< 0

4𝑥 − 8

1 − 𝑥< 0

𝑍𝑒𝑟𝑜: 2, 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑: 1 −∞, 1 : 𝑓 0 = −8 − 1, 2 : 𝑓 1.5 = 4 + 2,∞ : 𝑓 3 = −2 −

−∞,1 𝑈(2,∞)

6) Find compositions of functions (and domains).

Composition of Functions If 𝑓 𝑎𝑛𝑑 𝑔 are functions, then the composition of 𝑓 𝑎𝑛𝑑 𝑔 is

𝒇 ∘ 𝒈 𝒙 = 𝒇 𝒈 𝒙 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙 𝒊𝒏 𝒕𝒉𝒆 𝒅𝒐𝒎𝒂𝒊𝒏 𝒐𝒇 𝒈 𝒙

𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝒈 𝒙 𝒊𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒅𝒐𝒎𝒂𝒊𝒏 𝒐𝒇 𝒇(𝒙) • Check the domain of both the inner

function and the composition.

p. 348: 7

𝑓 𝑥 =𝑥 + 1

𝑥 − 1, 𝑔 𝑥 =

1

𝑥

(𝑓 𝑔 𝑥 =

1𝑥+ 1

1𝑥 − 1

=1 + 𝑥

1 − 𝑥

Domain: {x|𝑥 ≠ 0 𝑓𝑟𝑜𝑚 𝑔 , 𝑥 ≠ 1 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 }

7) Find the inverse of a function (and domains and ranges).

Finding Inverses

1) Replace 𝑓(𝑥) with 𝑦.

2) Interchange 𝑥 and 𝑦.

3) Solve for 𝑦.

4) Replace y with 𝑓−1 𝑥 .

• The domain of 𝑓 is the range of 𝑓−1.

• The domain of 𝑓−1 is the range of 𝑓.

p. 348: 11

𝑓 𝑥 =1

𝑥 − 1

𝑦 =1

𝑥 − 1

𝑥 =1

𝑦 − 1

𝑥𝑦 − 𝑥 = 1 𝑥𝑦 = 𝑥 + 1

𝑦 =𝑥 + 1

𝑥

𝑓−1 𝑥 =𝑥 + 1

𝑥

𝐷𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑓−1: 𝑥 𝑥 ≠ 0 𝑅𝑎𝑛𝑔𝑒 𝑜𝑓 𝑓−1: {𝑦|𝑦 ≠ 1}

8) Solve a logarithmic equation.

1) Isolate all logarithmic expressions on one side of the equation and use logarithmic properties to rewrite the logarithmic expressions as a single logarithmic expression whose coefficient is 1. 2) Rewrite the logarithmic expression as an exponential expression. 3) Solve. 4) Verify the solution. Reject values that produce the logarithm of a negative number or the logarithm of zero. Remember: the solution can be negative. It just can’t cause the logarithm of a negative number (or zero).

p. 349: 43

log6(𝑥 + 3) + log6(𝑥 + 4) = 1 log6(𝑥 + 3)(𝑥 + 4) = 1 61 = 𝑥 + 3 𝑥 + 4 𝑥2 + 7𝑥 + 12 = 6 𝑥2 + 7𝑥 + 6 = 0 𝑥 + 6 𝑥 + 1 = 0 𝑥 = −6, 𝑥 = −1 𝑥 ≠ −6 {−1}

9) Solve an exponential equation.

1) Isolate the exponential expression (if possible).

2) Take the common or natural logarithm of both sides.

3) Use the power rule: 𝑙𝑜𝑔𝑎𝑀𝑟 = 𝑟𝑙𝑜𝑔𝑎 𝑀

4) Solve for the variable.

p. 349: 38

5𝑥 = 3𝑥+2 ln 5𝑥 = ln 3𝑥+2 𝑥 ln 5 = 𝑥 + 2 ln 3 𝑥 ln 5 = 𝑥 ln 3 + 2 ln 3 𝑥 ln 5 − 𝑥 ln 3 = 2 ln 3 𝑥(ln 5 − ln 3) = 2 ln 3

𝑥 =2 ln 3

ln 5 − ln 3

≈ 4.301

10) Find the time required to reach a monetary goal.

Formulas:

𝐴 = 𝑃(1 +𝑟

𝑛)𝑛𝑡

𝐴 = 𝑃𝑒𝑟𝑡

p. 350: 50 (final part)

$10,000 𝑏𝑜𝑛𝑑 𝑓𝑢𝑛𝑑 4% 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑒𝑑 𝑠𝑒𝑚𝑖𝑎𝑛𝑛𝑢𝑎𝑙𝑙𝑦

How long to double?

𝐴 = 𝑃 1 +𝑟

𝑛

𝑛𝑡

= 10000 1 +0.04

2

2𝑡

20000 = 10000 1 +0.04

2

2𝑡

2 = 1.02 2𝑡 ln 2 = ln 1.022𝑡 ln 2 = 2𝑡 ln 1.02

𝑡 =ln 2

2 ln 1.02≈ 17.5 𝑦𝑒𝑎𝑟𝑠

11) Find the vertex, focus, and directrix, and orientation of opening of a parabola.

The parabola opens toward the focus and away from the directrix.

The focus and directrix are 𝑎 units from the vertex.

A 𝑦2 parabola opens right or left.

An 𝑥2 parabola opens up or down.

p. 698: 9

4𝑥2 − 16𝑥 + 16𝑦 + 32 = 0 4𝑥2 − 16𝑥 = −16𝑦 − 32 𝑥2 − 4𝑥 = −4𝑦 − 8

𝑥2 − 4𝑥 + 4 = −4𝑦 − 8 + 4 𝑥 − 2 2 = −4𝑦 − 4 𝑥 − 2 2 = −4 𝑦 + 1

𝑉𝑒𝑟𝑡𝑒𝑥 2, −1 𝑎 = 1

𝑂𝑝𝑒𝑛𝑠 𝑑𝑜𝑤𝑛 𝐹𝑜𝑐𝑢𝑠 2, −2 𝐷𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑥 𝑦 = 0

12) Write the equation of a hyperbola given the foci and the vertices.

𝑥2

𝑎2−𝑦2

𝑏2= 1 𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒 𝑎𝑥𝑖𝑠 𝑥

𝑦2

𝑎2−𝑥2

𝑏2= 1 (𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒 𝑎𝑥𝑖𝑠 𝑦)

𝑏2 = 𝑐2 − 𝑎2 𝐹𝑜𝑐𝑖: 𝑐, 𝑉𝑒𝑟𝑡𝑖𝑐𝑒𝑠: 𝑎

p. 699: 12

Hyperbola: center (0,0); focus 0,4 ; vertex (0, −2)

𝑇𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒 𝑎𝑥𝑖𝑠 𝑦 𝑐 = 4, 𝑎 = 2

𝑏2 = 𝑐2 − 𝑎2 = 42 − 22 = 12 𝑏2 = 12

𝑏 = 2 3 𝑦2

4−𝑥2

12= 1

13) Solve a determinant equation.

𝐷 = 𝑎 𝑏𝑐 𝑑

= 𝑎𝑑 − 𝑏𝑐

For a 3 X 3,remember to expand along one row or column. If you expand along row 1, the middle term will be multiplied by −1:

𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

= 𝑎11𝑎22 𝑎23𝑎32 𝑎33

− 𝑎12𝑎21 𝑎23𝑎31 𝑎33

+ 𝑎13𝑎21 𝑎22𝑎31 𝑎32

p. 741: 54

3 2 41 𝑥 50 1 −2

= 0

−1 2 3 −2𝑥 − 5 + −1 3 2 −2 + −1 4 4 1 = 0

−6𝑥 − 15 + 4 + 4 = 0 −6𝑥 − 7 = 0 −6𝑥 = 7

𝑥 = −7

6

14) Solve a curve fitting application.

Use your calculator to solve this problem.

1) Matrix Edit – enter the dimensions and then the entries of the augmented matrix of the system.

2) Matrix Math rref (option B on some calculators)

3) Matrix Names (choose A)

p. 795: 55

Find the quadratic function 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐

that passes through the points 0,1 , 1,0 , 𝑎𝑛𝑑 (−2,1)

1 = 𝑎 0 2 + 𝑏 0 + 𝑐 0 = 𝑎 1 2 + 𝑏 1 + 𝑐 1 = 𝑎 −2 2 + 𝑏 −2 + 𝑐

1 = 𝑎 0 2 + 𝑏 0 + 𝑐 0 = 𝑎 1 2 + 𝑏 1 + 𝑐 1 = 𝑎 −2 2 + 𝑏 −2 + 𝑐

𝑐 = 1

𝑎 + 𝑏 + 𝑐 = 0 4𝑎 − 2𝑏 + 𝑐 = 1

This can actually be worked as a system of two equations in two unknowns since c is known. Since the one on the test cannot be worked this way, let’s write an augmented matrix for a system of three equations.

0 0 11 1 14 −2 1

⋮⋮⋮

101

Place the matrix in the calculator and use the rref function. You will also have to convert the decimals to fractions using the FRAC option in MATH.

𝑎 = −1

3, 𝑏 = −

2

3, 𝑐 = 1

𝑦 = −1

3𝑥2 −2

3𝑥 + 1

15) Solve a system of nonlinear equations by any method.

Using substitution or elimination to solve this problem is probably the best idea.

p. 774: 82

The altitude of an isosceles triangle drawn to its base is 3 centimeters and its perimeter is 18 centimeters. Find the length of the base.

𝑥 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎 𝑠𝑖𝑑𝑒, 𝑦 =1

2𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒

𝑥 + 𝑥 + 𝑦 + 𝑦 = 18 𝑦2 + 32 = 𝑥2

x x

y y

𝑥 + 𝑥 + 𝑦 + 𝑦 = 18 𝑦2 + 32 = 𝑥2

2𝑥 + 2𝑦 = 18 𝑥 + 𝑦 = 9 𝑦 = 9 − 𝑥

𝑦2 + 32 = 𝑥2 9 − 𝑥 2 + 9 = 𝑥2

81 − 18𝑥 + 𝑥2 + 9 = 𝑥2 90 − 18𝑥 = 0 𝑥 = 5

𝑦 = 9 − 5 = 4

2 4 = 8 8 𝑐𝑒𝑛𝑡𝑖𝑚𝑒𝑡𝑒𝑟𝑠

16) Find the sum of a sequence.

𝑐

𝑛

𝑘=1

= 𝑐𝑛

𝑘

𝑛

𝑘=1

=𝑛(𝑛 + 1)

2

𝑘2𝑛

𝑘=1

=𝑛(𝑛 + 1)(2𝑛 + 1)

6

𝑘3𝑛

𝑘=1

=𝑛 𝑛 + 1

2

2

p. 838: 14

𝑘2 + 2 =

30

𝑘=1

𝑘230

𝑘=1

+ 2

30

𝑘=1

=30(30 + 1)(2(30) + 1)

6+ 2 30

= 9455 + 60 = 9515

17) Find a general formula for a given arithmetic sequence.

Two methods:

1) Set up a system of two equations in two unknowns.

OR

2) Find the slope. The slope is the common difference.

For both methods, you need the formula 𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑

p. 838: 21

10th term is 0, 18th term is 8 𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑 𝑎10 = 𝑎1 + 9𝑑 = 0 𝑎18 = 𝑎1 + 17𝑑 = 8

𝑎1 + 9𝑑 = 0 𝑎1 + 17𝑑 = 8

−1(𝑎1 + 9𝑑 = 0) 𝑎1 + 17𝑑 = 8

8𝑑 = 8 𝑑 = 1

𝑎1 + 9 1 = 0

𝑎1 = −9

𝑎𝑛 = −9 + 𝑛 − 1 1 = −9 + 𝑛 − 1 = 𝑛 − 10

10th term is 0, 18th term is 8

OR

𝑑 =8 − 0

18 − 10=8

8= 1

𝑎1 + 9 1 = 0

𝑎1 = −9

𝑎𝑛 = −9 + 𝑛 − 1 1 = −9 + 𝑛 − 1 = 𝑛 − 10

18) Work with applications of arithmetic sequences.

𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑

𝑆𝑛 = 𝑎1 + 𝑎2 +⋯+ 𝑎𝑛 =𝑛

2𝑎1 + 𝑎𝑛

p. 839: 34

A brick staircase has a total of 25 steps. The bottom step requires 80 bricks. Each successive step requires 3 fewer bricks. How many bricks are required for the top step and how many bricks are required to build the staircase?

𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑 𝑎25 = 80 + 24 −3 = 8

𝑆𝑛 = 𝑎1 + 𝑎2 +⋯+ 𝑎𝑛 =𝑛

2𝑎1 + 𝑎𝑛

𝑆25 =25

280 + 8 = 1100

19) Determine whether a geometric series converges or diverges (and determine the sum if it converges).

Geometric Series

Converges if 𝑟 < 1 with sum 𝑎1

1−𝑟

p. 838: 23

2 − 1 +1

2−1

4+⋯

𝑟 = −1 ÷ 2 = −1

2

𝐶𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠

𝑆𝑢𝑚 =𝑎11 − 𝑟=

2

1 − −12

=2

32

=4

3

20) Use the Binomial theorem to find a particular coefficient or term in a binomial expansion.

Binomial Theorem 𝑥 + 𝑐 𝑛

Start with 𝑛0𝑐0𝑥𝑛

The exponents of subsequent terms go up with the constant and down with the variable

p. 836: 30

What is the coefficient of 𝑥3 in the expansion of 𝑥 − 3 10?

The two exponents must add up to 10.

3 + 7 = 10

107−3 7𝑥3

120 ⋅ −2187 = −262440 −262,440