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**SBT Prize at the end of chapter 2 REAL NUMBERS (C2 Math 7 Horizon)**

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### Solve problem 1 page 45 SBT Math 7 Creative horizon episode 1

Express the following decimals as rational numbers: 12.3; 0.12; 5(3).

**Solution method**

We represent them with the denominator of 10,100,1000… or separate the integer and decimal parts to calculate.

**Detailed explanation**

\(\begin{array}{l}12.3 = \dfrac{{123}}{{10}};\\0.12 = \dfrac{{12}}{{100}} = \dfrac{3 }{{25}};\\5,(3) = 5 + 0,(3) = 5 + 3.0,(1) = 5 + 3.\dfrac{1}{9} = 5 + \dfrac{1 }{3} = \dfrac{{16}}{3}\end{array}\)

### Solution 2 page 45 SBT Math 7 Creative horizon episode 1

Please replace the sign ? by the appropriate number.

Sample. Since 32 = 9 \(\sqrt 9 \) = 3.

a) Since 42 = 16 \(\sqrt {16} \)= ?;

b) Since 92 = 81 \(\sqrt {81} \)= ?;

c) Since 12 = 1, \(\sqrt 1 \) = ?;

d) Since \({\left( {\dfrac{3}{5}} \right)^2}\)=\(\dfrac{9}{{25}}\) \(\sqrt {\dfrac {9}{{25}}} \)= ?

**Solution method**

We use the definition of the arithmetic square root to calculate and find the number to look for

**Detailed explanation**

a) Because 4^{2} = 16 so \(\sqrt {16} \) = 4;

b) Because 9^{2} = 81 so \(\sqrt {81} \) = 9;

c) Because 1^{2} = 1 so \(\sqrt 1 \)= 1;

d) Since \({\left( {\dfrac{3}{5}} \right)^2}\)=\(\dfrac{9}{{25}}\) \(\sqrt {\dfrac {9}{{25}}} \)=\(\dfrac{3}{5}\)

### Solve problem 3 pages 45 SBT Math 7 Creative horizon episode 1

Find the irrational number among the following: \(\sqrt 2 \);-\(\sqrt 4 \);\(\sqrt {\dfrac{{16}}{9}} \)

**Solution method**

We use the definition of irrational and rational numbers to distinguish.

**Detailed explanation**

We have: \(\sqrt 2 \) = 1.414213562… is an infinite non-recurring decimal number, so \(\sqrt 2 \) is an irrational number.

We have: 2^{2} = 4 so \(\sqrt 4 \)=2 so -\(\sqrt 4 \)=−2=\( – \dfrac{2}{1}\) is a rational number so -\(\sqrt 4 \) is a rational number.

We have: \({\left( {\dfrac{4}{3}} \right)^2} = \dfrac{{16}}{9}\) so \(\sqrt {\dfrac{{16} }{9}} \)=\(\dfrac{4}{3}\) is a rational number so \(\sqrt {\dfrac{{16}}{9}} \) is a rational number.

So only \(\sqrt 2 \) is an irrational number

### Solution 4 page 45 SBT Math 7 Creative horizon episode 1

Let’s calculate:

\(\sqrt {289} \);-\(\sqrt {144} \);\(\sqrt {\dfrac{{81}}{{225}}} \);\(\sqrt {{{( – 3)}^2}} \);\(\sqrt {{a^2}} \)

**Solution method**

We use the definition of exponentiation and square root of arithmetic to calculate

\(\sqrt{a^2}=|a|\)

+) If \(a \ge 0\) then \(|a|=a\)

+) If \(a < 0\) then \(|a|=-a\)

**Detailed explanation**

Because 17^{2} = 289 so \(\sqrt {289} \)=17;

Because 12^{2} = 144 so \(\sqrt {144} \)=12 or −\(\sqrt {144} \)=−12

Since \({\left( {\dfrac{9}{{15}}} \right)^2} = \dfrac{{81}}{{225}}\)so\(\sqrt {\dfrac{{ 81}}{{225}}} = \dfrac{9}{{15}}\)

We have: \(\sqrt {{{( – 3)}^2}} =|−3|=−(−3)=3\)

We have \(\sqrt {{a^2}} = \left| a \right|\)

+) When \(a \ge 0\) then \(|a|=a\)

+) When \(a < 0\) then \(|a|=-a\)

### Solve problem 5 pages 45 SBT Math 7 Creative horizon episode 1

Are the following statements true or false? If false, re-state it correctly:

a) \(\sqrt {36} \) Q

b) \(\sqrt 7 \) R

c) 0.23 \( \notin \) CHEAP

d) \( – \sqrt 3 \) R

**Solution method**

First we have to find the roots and then use the definition of irrational and rational sets to find true and false statements.

**Detailed explanation**

a) We have 6^{2} = 36 so \(\sqrt {36} \)=6 is a rational number deduced \(\sqrt {36} \)∈Q. Hence a) is correct.

b) We have: \(\sqrt 7 \) = 2.645751311 is an infinite non-cyclic decimal number, so \(\sqrt 7 \) is an irrational number, and an irrational number is a deduced real number \(\ sqrt 7 \) ∈R. Hence b) is correct.

c) We have: 0.23 = \(\dfrac{{23}}{{100}}\) (where 23; 100 ∈ ℤ and 100 ≠ 0) is a rational number where the rational number is an interpolated real number. out 0.23∈R. Hence c) is false.

d) We have: \( – \sqrt 3 \) = -1.7320508075… is an infinite non-recurring decimal number, so \( – \sqrt 3 \) is an irrational number, and an irrational number is an interpolated real number. out \( – \sqrt 3 \)∈R. Hence d) is true

### Solve lesson 6 page 45 SBT Math 7 Creative horizon episode 1

Find x, know: \({\left( {x + 9} \right)^2} = 5\;\)

**Solution method**

\(A^2=b \Leftrightarrow A=\sqrt{b}\) or \(A=-\sqrt{b}\)

**Detailed explanation**

(x + 9)^{2} = 5

x + 9 = \(\sqrt 5 \) or x + 9 = −\(\sqrt 5 \)

+) Case 1: x + 9 = \(\sqrt 5 \)

x = \(\sqrt 5 \) – 9

+) Case 2: x + 9 = −\(\sqrt 5 \)

x = −\(\sqrt 5 \)– 9

So x = \(\sqrt 5 \) – 9 or x = −\(\sqrt 5 \) – 9

### Solve problem 7 page 45 SBT Math 7 Creative horizon episode 1

As of April 25, 2019, Hanoi has a total population of 8 053 663 people, including 3 991 919 men and 4 061 744 women (source: https://hanoimoi.com.vn/tintuc/xahoi/) . Let’s round the numbers to hundreds.

**Solution method**

adsense

To round decimals to a certain rounding row:

Step 1: Underline the decimal place of the rounded row.

Step 2: Look at the digit immediately to the right

+ If the digit is greater than or equal to 5, increase the underscore by 1 and then replace all the digits to the right with 0 or omit them if they are in decimal.

+ If the digit is less than 5, keep the dashed digit and replace all the digits to the right with 0 or omit it if they are in the decimal part.

**Detailed explanation**

The total population of Hanoi is 8 053 663, this number has 6 digit of the rounded row, the number after the rounded row is 6 > 5, so we add 1 unit to the rounded row digit, the following digits Rounded row is replaced by 0, we get:

8 053 663 ≈ 8 053 700.

The total population of Hanoi male is: 3 991 919, this number has the number of rounded row digits 9, the number after the rounded row is 1 < 5, so we keep the number of rounded rows and letters. The number after the rounding row is replaced by zero, we get:

3 991 919 ≈ 3 991 900 .

The total population of Hanoi male is: 4 061 744, this number has the rounded digit of the row, the number after the rounded row is 4 < 5, so we keep the number of the rounded row and the letters. The number after the rounding row is replaced by zero, we get:

4 061 744 4 061 700

### Solve problem 8 page 46 SBT Math 7 Creative horizon episode 1

Calculate the value rounded to the units of the expression \(A = \dfrac{{99,21.5,89}}{{3,05}}\) in two ways as follows:

Method 1. Round the number before performing the calculation.

Method 2. Do the calculation first and then round the number.

**Solution method**

With method 1, we will round the numbers to the units of both numerator and denominator and then calculate .

With method 2, we will perform the calculation first after the decimal number and then reduce the quotient.

**Detailed explanation**

Method 1:

We have: 99.21 ≈ 99; 5.89 ≈ 6; 3.05 3

Then: \(A = \dfrac{{99,21.5,89}}{{3.05}} \approx \dfrac{{99.6}}{3} = 198\)

So the value of A is approximately 198 .

Method 2:

\(A = \dfrac{{99,21.5,89}}{{3.05}} = \dfrac{{584.3469}}{{3.05}} = 191,589… \approx 192\)

So by way of number 1 we can calculate A 198 and by way of number 2 we can calculate A 192

### Solve problem 9 page 46 SBT Math 7 Creative horizon episode 1

The results of your Literature subject Autumn in semester 2 are as follows:

Factor 1; 5; 8;

Factor 2; 7; 9;

Factor 3; 7.

Calculate Thu’s Literature grade point average and round to tenths.

**Solution method**

We calculate your total score Thu with all the coefficients then divide it equally by the total score multiplied by the coefficient to find the average score of your Literature subject.

**Detailed explanation**

Thu’s average score in Literature is:

\(\dfrac{{5 + 8 + 7.2 + 9.2 + 7.3}}{9} = \dfrac{{66}}{9} = 7,33333… \approx 7.3\)

So, Thu’s average score in Literature in semester 2 is 7.3.

### Solve problems 10 pages 46 SBT Math 7 Creative horizon episode 1

Round the following numbers to the hundreds: 3000π; – 200\(\sqrt 3 \)

**Solution method**

First we write the numbers \(\pi ;\sqrt 3 \) as decimals (just take and round to units) then multiply the found number by 3000 and -200 and then round to hundreds

**Detailed explanation**

We have: \(3000\pi = 9424,7777961…\)

This number has a rounded digit of 4, the digit after the rounded row is 2 < 5, so we keep the number of the rounded row, the digits of the units and tens are replaced by zeros, the decimal digits Discarding, we get:

\(3000\pi\) = 9424,7777961… ≈ 9400.

We have: – 200\(\sqrt 3 \) = – 346,4101615…

This number has the rounded digit of 3, the digit after the rounded row is 4 < 5, so we keep the rounded digit, the digits of the units and tens are replaced by 0, the decimal places. Discarding, we get:

– 200\(\sqrt 3 \) = – 346,4101615… ≈ – 300

### Solve problems 11 pages 46 SBT Math 7 Creative horizon episode 1

Use a calculator to calculate and round the following numbers to the nearest hundredth: −250\(\sqrt 3 \); π\(\sqrt 2 \);\(\sqrt {13} – \sqrt 5 \)

**Solution method**

I use a calculator to do the math

**Detailed explanation**

We have: −250\(\sqrt 3 \)= – 433.0127019…

This number has the rounded row digit as 1, the digit after the rounded row is 2 < 5, so we keep the rounded row digit, the decimal places after the rounding row are removed, we get:

−250\(\sqrt 3 \)= – 433.0127019… ≈ – 433.01

We have: \(pi\)\(\sqrt 2 \)= 4.442882938…

This number has the rounded row digit of 4, the digit after the rounded row is 2 < 5, so we keep the rounded row digit, the decimal places after the rounding row are removed, we get:

\(pi\)\(\sqrt 2 \)= 4.442882938… ≈ 4.44.

We have: \(\sqrt {13} – \sqrt 5 \) = 1.369483298…

This number has the rounded row digit of 6, the digit after the rounded row is 9 > 5, so we add 1 unit to the rounded row digit, the decimal places after the rounding row are removed, we get :

\(\sqrt {13} – \sqrt 5 \)= 1.369483298… ≈ 1.37

### Solve problem 12 pages 46 SBT Math 7 Creative horizon episode 1

Calculate the perimeter and area of a rectangular garden with a length of 15.24m and a width of 9.4m and then round to units.

**Solution method**

We calculate the perimeter and area of the garden and then round the area and perimeter to units.

**Detailed explanation**

Perimeter of a rectangular garden is:

2(15.24 + 9.4) = 49.28 ≈ 49 (m).

The area of a rectangular garden is:

15,24.9.4 = 143,256 143 (m^{2}).

So the perimeter and area of a rectangular garden and rounded to units are 49m and 143m respectively.^{2 }