Articles

1.5E: Exercises - Mathematics


Practice Makes Perfect

Multiply Integers

In the following exercises, multiply.

Exercise (PageIndex{55})

(−4cdot 8)

Answer

-32

Exercise (PageIndex{56})

(-3cdot 9)

Exercise (PageIndex{57})

(9(-7))

Answer

-63

Exercise (PageIndex{58})

(13(-5))

Exercise (PageIndex{59})

(-1cdot 6)

Answer

-6

Exercise (PageIndex{60})

(-1cdot 3)

Exercise (PageIndex{61})

(-1(-14))

Answer

14

Exercise (PageIndex{62})

(-1(-19))

Divide Integers

In the following exercises, divide.

Exercise (PageIndex{63})

(-24div 6)

Answer

-4

Exercise (PageIndex{64})

(35div (-7))

Exercise (PageIndex{65})

(-52 div (-4))

Answer

13

Exercise (PageIndex{66})

(-84 div (-6))

Exercise (PageIndex{67})

(-180 div 15)

Answer

-12

Exercise (PageIndex{68})

(-192div 12)

Simplify Expressions with Integers

In the following exercises, simplify each expression.

Exercise (PageIndex{69})

5(−6)+7(−2)−3

Answer

-47

Exercise (PageIndex{70})

8(−4)+5(−4)−6

Exercise (PageIndex{71})

((-2)^{6})

Answer

64

Exercise (PageIndex{72})

((-3)^{5})

Exercise (PageIndex{73})

((-4)^{2})

Answer

-16

Exercise (PageIndex{74})

((-6)^{2})​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​

Exercise (PageIndex{75})

−3(−5)(6)

Answer

90

Exercise (PageIndex{76})

−4(−6)(3)

Exercise (PageIndex{77})

(8−11)(9−12)

Answer

9

Exercise (PageIndex{78})

(6−11)(8−13)

Exercise (PageIndex{79})

26−3(2−7)

Answer

41

Exercise (PageIndex{80})

23−2(4−6)

Exercise (PageIndex{81})

(65div (−5)+(−28)div (−7))

Answer

-9

Exercise (PageIndex{82})

(52div(−4)+(−32)div(−8))

Exercise (PageIndex{83})

9−2[3−8(−2)]

Answer

-29

Exercise (PageIndex{84})

11−3[7−4(−20)]

Exercise (PageIndex{85})

((−3)^{2}−24div (8−2))

Answer

5

Exercise (PageIndex{86})

((−4)^{2}−32div (12−4))

Evaluate Variable Expressions with Integers

In the following exercises, evaluate each expression.

Exercise (PageIndex{87})

y+(−14) when

  1. y=−33
  2. y=30
Answer
  1. −47
  2. 16

Exercise (PageIndex{88})

x+(−21) when

  1. x=−27
  2. x=44

Exercise (PageIndex{89})

  1. a+3 when a=−7
  2. −a+3 when a=−7
Answer
  1. −4
  2. 10

Exercise (PageIndex{90})

  1. d+(−9) when d=−8
  2. −d+(−9) when d=−8

Exercise (PageIndex{91})

m+n when
m=−15,n=7

Answer

-8

Exercise (PageIndex{92})

p+q when
p=−9,q=17

Exercise (PageIndex{93})

r+s when r=−9,s=−7

Answer

-16

Exercise (PageIndex{94})

t+u when t=−6,u=−5

Exercise (PageIndex{95})

((x+y)^{2}) when
x=−3,y=14

Answer

121

Exercise (PageIndex{96})

((y+z)^{2}) when
y=−3, z=15

Exercise (PageIndex{97})

−2x+17 when

  1. x=8
  2. x=−8
Answer
  1. 1
  2. 33

Exercise (PageIndex{98})

−5y+14 when

  1. y=9
  2. y=−9

Exercise (PageIndex{99})

10−3m when

  1. m=5
  2. m=−5
Answer
  1. −5
  2. 25

Exercise (PageIndex{100})

18−4n when

  1. n=3
  2. n=−3

Exercise (PageIndex{101})

(2w^{2}−3w+7) when
w=−2

Answer

21

Exercise (PageIndex{102})

(3u^{2}−4u+5)

Exercise (PageIndex{103})

9a−2b−8 when
a=−6 and b=−3

Answer

-56

Exercise (PageIndex{104})

7m−4n−2 when
m=−4 and n=−9

​​​​​Translate English Phrases to Algebraic Expressions

In the following exercises, translate to an algebraic expression and simplify if possible.

Exercise (PageIndex{105})

the sum of 3 and −15, increased by 7

Answer

(3+(−15))+7;−5

Exercise (PageIndex{106})

the sum of −8 and −9, increased by 23

Exercise (PageIndex{107})

the difference of 10 and −18

Answer

10−(−18);28

Exercise (PageIndex{108})

subtract 11 from −25

Exercise (PageIndex{109})

the difference of −5 and −30

Answer

−5−(−30);25

Exercise (PageIndex{110})

subtract −6 from −13

Exercise (PageIndex{111})

the product of −3 and 15

Answer

(−3cdot 15);−45

Exercise (PageIndex{112})

the product of −4 and 16

Exercise (PageIndex{113})

the quotient of −60 and −20

Answer

(−60div(−20));3

Exercise (PageIndex{114})

the quotient of −40 and −20

Exercise (PageIndex{115})

the quotient of −6 and the sum of a and b

Answer

(frac{-6}{a + b})

Exercise (PageIndex{116})

the quotient of −6 and the sum of a and b

Exercise (PageIndex{117})

the product of −10 and the difference of p and q

Answer

−10(p−q)

Exercise (PageIndex{118})

the product of −13 and the difference of c and d

​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Use Integers in Applications

In the following exercises, solve.

Exercise (PageIndex{119})

Temperature On January 15, the high temperature in Anaheim, California, was 84°. That same day, the high temperature in Embarrass, Minnesota was −12°. What was the difference between the temperature in Anaheim and the temperature in Embarrass?

Answer

96°

Exercise (PageIndex{120})

Temperature On January 21, the high temperature in Palm Springs, California, was 89°, and the high temperature in Whitefield, New Hampshire was −31°. What was the difference between the temperature in Palm Springs and the temperature in Whitefield?

Exercise (PageIndex{121})

Football At the first down, the Chargers had the ball on their 25 yard line. On the next three downs, they lost 6 yards, gained 10 yards, and lost 8 yards. What was the yard line at the end of the fourth down?

Answer

21

Exercise (PageIndex{122})

Football At the first down, the Steelers had the ball on their 30 yard line. On the next three downs, they gained 9 yards, lost 14 yards, and lost 2 yards. What was the yard line at the end of the fourth down?

Exercise (PageIndex{123})

Checking Account Mayra has $124 in her checking account. She writes a check for $152. What is the new balance in her checking account?

Answer​​​​​​​

−$28

Exercise (PageIndex{124})

Checking Account Selina has $165 in her checking account. She writes a check for $207. What is the new balance in her checking account?

Exercise (PageIndex{125})

Checking Account Diontre has a balance of −$38 in his checking account. He deposits $225 to the account. What is the new balance?

Answer

$187

Exercise (PageIndex{126})

Checking Account Reymonte has a balance of −$49 in his checking account. He deposits $281 to the account. What is the new balance?

Everyday Math

​​​​​​​

Exercise (PageIndex{127})

Stock market Javier owns 300 shares of stock in one company. On Tuesday, the stock price dropped $12 per share. What was the total effect on Javier’s portfolio?

Answer

Weight loss In the first week of a diet program, eight women lost an average of 3 pounds each. What was the total weight change for the eight women?

Exercise (PageIndex{128})

Weight loss In the first week of a diet program, eight women lost an average of 3 pounds each. What was the total weight change for the eight women?

Writing Exercises

Exercise (PageIndex{129})

In your own words, state the rules for multiplying integers.

Answer

Answers may vary

Exercise (PageIndex{130})

In your own words, state the rules for dividing integers.

Exercise (PageIndex{131})

Why is (−2^{4} eq (−2)^{4})?

Answer

Answers may vary

Exercise (PageIndex{132})

Why is (−4^{3} eq (−4)^{3})?​​​​​​

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

​​​​​​​


1.5E: Exercises - Mathematics

Here is another geometric application of the integral: find the length of a portion of a curve. As usual, we need to think about how we might approximate the length, and turn the approximation into an integral.

We already know how to compute one simple arc length, that of a line segment. If the endpoints are $ds P_0(x_0,y_0)$ and $ds P_1(x_1,y_1)$ then the length of the segment is the distance between the points, $ds sqrt<(x_1-x_0)^2+(y_1-y_0)^2>$, from the Pythagorean theorem, as illustrated in figure 11.4.1.

Now if the graph of $f$ is "nice'' (say, differentiable) it appears that we can approximate the length of a portion of the curve with line segments, and that as the number of segments increases, and their lengths decrease, the sum of the lengths of the line segments will approach the true arc length see figure 11.4.2.

Now we need to write a formula for the sum of the lengths of the line segments, in a form that we know becomes an integral in the limit. So we suppose we have divided the interval $[a,b]$ into $n$ subintervals as usual, each with length $Delta x =(b-a)/n$, and endpoints $ds a=x_0$, $ds x_1$, $ds x_2$, &hellip, $ds x_n=b$. The length of a typical line segment, joining $ds (x_i,f(x_i))$ to $ds (x_,f(x_))$, is $dssqrt<(Delta x )^2 +(f(x_)-f(x_i))^2>$. By the Mean Value Theorem (6.5.2), there is a number $ds t_i$ in $ds (x_i,x_)$ such that $ds f'(t_i)Delta x=f(x_)-f(x_i)$, so the length of the line segment can be written as $ sqrt<(Delta x)^2 + (f'(t_i))^2Delta x^2>= sqrt<1+(f'(t_i))^2>,Delta x. $ The arc length is then $ lim_sum_^ sqrt<1+(f'(t_i))^2>,Delta x= int_a^b sqrt<1+(f'(x))^2>,dx. $ Note that the sum looks a bit different than others we have encountered, because the approximation contains a $ds t_i$ instead of an $ds x_i$. In the past we have always used left endpoints (namely, $ds x_i$) to get a representative value of $f$ on $ds [x_i,x_]$ now we are using a different point, but the principle is the same.

To summarize, to compute the length of a curve on the interval $[a,b]$, we compute the integral $int_a^b sqrt<1+(f'(x))^2>,dx.$ Unfortunately, integrals of this form are typically difficult or impossible to compute exactly, because usually none of our methods for finding antiderivatives will work. In practice this means that the integral will usually have to be approximated.

Example 11.4.1 Let $ds f(x) = sqrt$, the upper half circle of radius $r$. The length of this curve is half the circumference, namely $pi r$. Let's compute this with the arc length formula. The derivative $f'$ is $ds ds -x/sqrt$ so the integral is $ int_<-r>^r sqrt<1+>,dx =int_<-r>^r sqrt,dx =rint_<-r>^r sqrt<1over r^2-x^2>,dx. $ Using a trigonometric substitution, we find the antiderivative, namely $ds arcsin(x/r)$. Notice that the integral is improper at both endpoints, as the function $ds sqrt<1/(r^2-x^2)>$ is undefined when $x=pm r$. So we need to compute $ lim_int_D^0 sqrt<1over r^2-x^2>,dx + lim_int_0^D sqrt<1over r^2-x^2>,dx. $ This is not difficult, and has value $pi$, so the original integral, with the extra $r$ in front, has value $pi r$ as expected.


Larson Algebra 2 Solutions Chapter 12 Probability and Statistics Exercise 12.1

Chapter 12 Probability and Statistics Exercise 12.1 1E

Chapter 12 Probability and Statistics Exercise 12.1 1GP

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Four Strategies to Help Struggling Math Students

I have heard a million excuses from my students when asked why they have a bad grade in math. It’s too boring. I don’t get it. The teacher hates me. Whatever the reason, there are some strategies teachers can use in the math classroom to better reach students who may have thrown in the towel. I have compiled a list of what’s worked for some of my Math RTI students, and here they are:

Use the flipped classroom

The number one reason my students say they don’t “get” math is because the teacher talks too fast. I hear it all the time . They tell me when their math teacher asks, “any questions?” They haven’t even digested the information enough to formulate a question, and then. once they do have a question, they’ve moved on to something new.

If you are unfamiliar with this teaching strategy, The Flipped Classroom involves recording your math lesson and making the videos available online for students to watch at home, or outside of class time. With the flipped classroom, struggling students can pause the video, think about what is being said, try it in their notes, and press play when they are ready to do so. They can watch, pause, re watch, as many times as they individually need. Some teachers require students to take notes, or to work out some of the problems as they watch the video. This way, when students come to class, class time is used to practice the lesson, with the teacher acting as a coach, conferencing with students, correcting misconceptions, and providing support or extensions for students who need it. Even if the idea of a “flipped classroom” scares you, recording your lecture is never a bad idea for your struggling students.

Students who struggle in math need time. The flipped classroom is the perfect solution for the math student who works slowly . One of the reasons I love this teaching strategy is because my students who struggle in math need to see you solve a problem, not once, not twice, not even three times- they need to see it and try it sometimes as many as ten times! (I am not even exaggerating, I’ve witnessed it firsthand).

Allow students to have an example near them while they work, yes- even on assessments

In my experience, if a student is struggling in math, they may also struggle with their working memory. This, according to Dr. Matthew Kruger from ADDitude magazine, can “impair a child’s ability to follow multi-step directions, tap into old information, or quickly recall lessons.” All the time, I have students who say “I could do this yesterday, but as soon as I got the quiz I couldn’t remember what to do!” If they have an example to look at, it helps them remember the steps they’ve been practicing. Once the student feels confident, they will gradually need these less and less.

Encourage them to “think out loud” when solving math problems

Having my students talk through their problems is something I practice daily. Sometimes if I hear a really great explanation, I will stop the class so everyone can hear it (of course they roll their eyes when I do this, but if its a great explanation they have to share!) Research shows that students learn better when they explain their own problem-solving steps ( Berardi-Coletta, Buyer, Dominowsky, & Rellinger, 1995 ). This may take practice, and might be uncomfortable for students at first, but you can help by constantly modeling your thinking and asking questions like “How did you get that answer?” or “Why did you choose that operation?” This allows them to develop a deeper understanding of what they are doing, and it also helps me to better understand what they are thinking, so I can address any misconceptions.

Frequently check their work, and give immediate feedback

The reason I love teaching math, as opposed to reading, is you can clearly see a student’s mistakes. Because of this, it is fairly easy to see where a student is struggling. Sometimes its computation, sometimes they use the wrong operation, sometimes they skip a step or use a formula incorrectly. Regardless of the error, you can spot it right away and correct it with them - IF you catch it. Students who struggle need more of your time and attention. Make a point to stop by their desks while they are working and check on their work. If they aren’t getting it, they need more examples, more support, more guidance. If you are lucky enough to have an aid in your class, have them monitor the students who are getting it, and pull a group of struggling students aside and really dissect what mistakes they are making, and address them.

Berardi-Coletta, B., Buyer, L. S., Dominowski, R. L., & Rellinger, E. R. (1995). Metacognition and problem solving: A process-oriented approach. Journal of Experimental Psychology: Learning, Memory, and Cognition , 21, 205–223.

Cruger, Matthew. "15 Memory Exercises for Forgetful Kids." ADDitude . New Hope Media LLC, 15 Dec. 2016. Web. 20 June 2017.


Larson Algebra 2 Solutions Chapter 13 Trigonometric Ratios and Functions Exercise 13.1

Chapter 13 Trigonometric Ratios and Functions Exercise 13.1 1E

Chapter 13 Trigonometric Ratios and Functions Exercise 13.1 1GP

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1.5E: Exercises - Mathematics

In Exercises 33 - 38, evaluate the function at the indicated value of $ x $. Round your result to three decimal places.

Answer

More Answers

Topics

Exponential and Logarithmic Functions

Precalculus with Limits

Exponential and Logarithmic Functions

Exponential Functions and Their Graphs

Discussion

Can you show your work please

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Video Transcript

Okay, here we have function f of X, and we're going to find f of 240 by substituting 240 for X and then using a calculator. So we're computing 1.5 times each of the 120. So then in the calculator we can take the 1.5. And then if you need to know where to find E, look next to your four to the left of before there's a button that says Natural log on. The second function of that is E to the X to repress second natural log and we get E to a power and then we're typing in 1 20 So notice the answer is 1.956 e 52. That means the number was so large they had to put it into scientific notation. So this is 1.956 times 10 to the 52nd


5 Answers 5

It's not ugly, but exactly what's expected. If you type

then you expect that there is some space around the minus sign, because it denotes an operation. When you type $1e-10$ , TeX interprets it in exactly the same way, because it can't read your mind: the two expressions are formally the same, only two symbols are different.

If you want that an expression that's normally interpreted as a polynomial should be treated in a different way, then you have to properly mark it.

because in this case the braces around -10 tell TeX to enter a subformula and so the minus sign is initial, so not interpreted as a binary operation, but as a unary operator.

You could make a definition, such as

but there's a much better alternative, the package siunitx .

This package offers many more features than just printing numbers in the desired format consult its documentation to find them.

Note that siunitx is not understood by MathJax, so with it you must stick to the “hand made” solution. You can still say, in it,


The Math.sqrt method returns the positive square root of the argument given to it. The method will take a numerical value of any type as argument. It returns a value of type double. If the method is given a negative value, it will return the value NaN.

  1. Math.pow(8,1.0/3.0) => 2.0
  2. Math.pow(2,10) => 1024.0
  3. Math.pow(-32.0,0.2) => NaN
  4. (int)Math.pow(100,0.25) => 3

Exercises 16 (Plotting in rectangular coordinates)

Show that the stress is related to the strain by a law of the form (sigma=Evarepsilon) , where (E) is a constant. Determine the law for the wire under test.

  1. During a test on a simple lifting machine, the following results were obtained showing the applied force, (F) , for the load, (L) , lifted:

It is thought that the equation relating (F) and (L) is of the form (F = kL + c) where both (k) and (c) are constants. Assuming that the law holds true, find the force necessary to lift a load of (1,mathrm) .

  1. The variation in pressure, (p) , within a vessel at a temperature, (T) , follows a law of the form (p = aT + b) . Verify that the data below relates the data by this law and determine the law.

Determine graphically the solution to the simultaneous equations: [egin 2.5x + 0.45 - 3y &= 0 1.6x + 0.8y - 0.8 &= 0 end]

Plot the graph of (y = 4x^3 - 4x^2 - 15x + 18) for values of (x) from (-3) to (+3) and using the graph, determine the roots of the polynomial.

On the same axes and to the same scale plot the equations (y = 1.5e^<-1.18x>) and (y = 1.1(1 - e^<-2.3x>)) . Determine the solution to the equations from your graph.


Application & Effectiveness

The 5E Model is most effective when students are encountering new concepts for the very first time because there is opportunity for a complete learning cycle.

According to co-creator Rodger W. Bybee, the 5E Model is best used in a unit of two to three weeks in which each phase is the basis for one or more distinct lessons. “Using the 5Es model as the basis for a single lesson decreases the effectiveness of the individual phases due to shortening the time and opportunities for challenging and restructuring of concepts and abilities—for learning,” Bybee explains. And if too much time is spent on each phase, the structure isn’t as effective and students may forget what they’ve learned.

The following research findings illustrate the positive impact of the 5E Model in classrooms:

One study showed that the 5E Model caused “a significantly better acquisition of scientific conceptions…than traditional instruction,” according to Biochemistry and Molecular Biology Education.

One study found that the 5E Instructional Model significantly increased learning and retention of science lessons.

The International Journal on New Trends in Education and Their Implications found the 5E learning cycle model positively affects student achievement and the permanence of knowledge.

The 5E Model allows educators to create a unique learning experience for students. Teachers who can incorporate instructional models like the 5E Model into their classrooms help students build a strong foundation of knowledge through active participation.

Lesley University’s online Master of Education programs equip teachers with the knowledge and tools to effectively educate students in the modern classroom. With specialized degrees in mathematics education, science in education, and more, Lesley offers opportunities for educators to deepen their understanding of current approaches and hone their teaching skills and assessment strategies.