## SBT Prize at the end of chapter 2 (C2 Math 7 Horizon) – Math Book

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 42 = 16 so $$\sqrt {16}$$ = 4;

b) Because 92 = 81 so $$\sqrt {81}$$ = 9;

c) Because 12 = 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: 22 = 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 172 = 289 so $$\sqrt {289}$$=17;

Because 122 = 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 62 = 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

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 (m2).

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