#### Maintaining Mathematical Proficiency

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##### Sections
###### Exercises
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Exercises 1 According to the order of operations, we must remember to evaluate the exponent first, the division and multiplication next, while addition and subtraction are evaluated last. Let's do it for the given expression!OperationBefore SimplificationAfter Simplification Calculate power12(214​)−33+15−9212(214​)−27+15−81 Calculate quotient12(214​)−27+15−8112(7)−27+15−81 Multiply12(7)−27+15−8184−27+15−81 Subtract term84−27+15−8157+15−81 Add terms57+15−8172−81 Subtract term72−81-9The expression equals -9. Alternative solution info Using a Calculator We can also evaluate this expression using a calculator. To do so, type it as shown below.
Exercises 2 According to the order of operations, we must remember to evaluate the exponent first, the division and multiplication next, while addition and subtraction are evaluated last. Let's do it for the given expression!OperationBefore SimplificationAfter Simplification Calculate power52⋅8÷22+20⋅3−425⋅8÷4+20⋅3−4 Multiply25⋅8÷4+20⋅3−4200÷4+60−4 Divide by 4200÷4+60−450+60−4 Add and subtract terms50+60−4106The expression equals 106. Alternative solution info Using a Calculator We can also evaluate this expression using a calculator. To do so, type it as shown below.
Exercises 3 According to the order of operations, expressions inside parentheses are evaluated first, while addition and subtraction are evaluated last. For the given expression, this means evaluating the expression inside the brackets completely before adding the product. -7+16÷24+(10−42)​ Notice also that even within this parenthetical expression, we must obey the order of operations. 10−42​ For this, we must remember to evaluate the exponent first, the multiplication next, and finally the difference between their results.OperationBefore SimplificationAfter Simplification Calculate power-7+16÷24+(10−42)-7+16÷24+(10−16) Subtract term-7+16÷24+(10−16)-7+16÷24+(-6) Calculate power-7+16÷24+(-6)-7+16÷16+(-6) Divide by 16-7+16÷16+(-6)-7+1+(-6) Add and subtract terms-7+1+(-6)-12The expression equals -12. Alternative solution info Using a Calculator We can also evaluate this expression using a calculator. To do so, type it as shown below.
Exercises 4 When we are given a number in a radical, we are typically being asked for the principal root. When the index is even, this is the positive root. Since the index of 64​ is even, let's find its principal root. 64​Split into factors8⋅8​a⋅a=a282​a2​=a8 The real root of 64​ is 8.
Exercises 5 When we are given a number in a radical, we are typically being asked for the principal root. When the index is even, this is the positive root. Since the index of -4​ is even, let's find its principal root. -4​Split into factors-2⋅2​a⋅a=a2-22​a2​=a-2 The real root of -4​ is -2.
Exercises 6 When we are given a number in a radical, we are typically being asked for the principal root. When the index is even, this is the positive root. Since the index of -25​ is even, let's find its principal root. -25​Split into factors-5⋅5​a⋅a=a2-52​a2​=a-5 The real root of -25​ is -5.
Exercises 7 When we are given a number in a radical, we are typically being asked for the principal root. When the index is even, this is the positive root. Since the index of ±121​ is even, let's find its principal root. ±121​Split into factors±11⋅11​a⋅a=a2±112​a2​=a±11 The real root of ±121​ is ±11.
Exercises 8 Explicit equations for arithmetic sequences follow a specific format. an​=a1​+(n−1)d​ In this form, a1​ is the first term in a given sequence, d is the common difference from one term to the next, and an​ is the nth term in the sequence. For this exercise, the first term is a1​=12. Let's observe the other terms to determine the common difference d. 12→+214→+216→+218…​ By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an​=a1​+(n−1)dd=2, a1​=12an​=12+(n−1)(2)Distribute 2an​=12+2n−2Subtract terman​=2n+10 The equation for the nth term of the given arthmetic sequence is an​=2n+10.
Exercises 9 Explicit equations for arithmetic sequences follow a specific format. an​=a1​+(n−1)d​ In this form, a1​ is the first term in a given sequence, d is the common difference from one term to the next, and an​ is the nth term in the sequence. For this exercise, the first term is a1​=6. Let's observe the other terms to determine the common difference d. 6→-33→-30→-3-3…​ By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an​=a1​+(n−1)dd=-3, a1​=6an​=6+(n−1)(-3)Distribute -3an​=6−3n+3Add termsan​=-3n+9 The equation for the nth term of the given arthmetic sequence is an​=-3n+9.
Exercises 10 Explicit equations for arithmetic sequences follow a specific format. an​=a1​+(n−1)d​ In this form, a1​ is the first term in a given sequence, d is the common difference from one term to the next, and an​ is the nth term in the sequence. For this exercise, the first term is a1​=22. Let's observe the other terms to determine the common difference d. 22→-715→-78→-71…​ By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an​=a1​+(n−1)dd=-7, a1​=22an​=22+(n−1)(-7)Distribute -7an​=22−7n+7Add termsan​=-7n+29 The equation for the nth term of the given arthmetic sequence is an​=-7n+29.
Exercises 11 For a number to be a perfect square , its square root must be an integer. This means that if an integer A is a perfect square it can be written as A=a2, where a is another integer. ​A​=a2​A​=a​ Let's consider the product of two perfect squares. AB=a2b2​ For simplicity we are considering that A, B, a, and b are positive integers. However, the result we will obtain is valid for all integer values. For this product to be a perfect square, its square root must be an integer. Let's check if this is the case. A⋅B​A=a2, B=b2a2⋅b2​a2=a⋅aa⋅a⋅b⋅b​Commutative Property of Multiplicationa⋅b⋅a⋅b​Associative Property of Multiplication(ab)⋅(ab)​a⋅a=a2(ab)2​a2​=a(ab)2​a2​=aab Since a and b are integers, so is their product ab. Therefore, we have shown that the square root of the product of two perfect squares, A⋅B​, is always an integer. Consequently, by definition, the product of two perfect squares is always a perfect square. Let's now can consider the quotient of two perfect squares. BA​=b2a2​​ Once more, A, B, a, and b are all positive integers, and b​=0. For this quotient to be a perfect square, its square root must be an integer. Let's check if this is the case. BA​​A=a2, B=b2b2a2​​a2=a⋅ab⋅ba⋅a​​Rewrite b⋅ba⋅a​ as ba​⋅ba​ba​⋅ba​​a⋅a=a2(ba​)2​a2​=aba​ As we can see, the square root of the quotient of two perfect squares is equal to the quotient of two integers. However, this isn't always an integer. Consider the example shown below. 21​=0.5​ Therefore, since the square root of the quotient of two perfect squares is not always an integer, by definition, the quotient of two perfect squares is not always a perfect square.