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Solving SBT Lesson 1: Algebraic Expressions (C7 SBT Math 7 Horizons)
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Solution 1 page 25 SBT Math 7 Creative horizon episode 2 – CTST
Write a numerical expression representing the area of a parallelogram whose base length is 6 cm and height is 5 cm.
Detailed instructions for solving Lesson 1
Solution method
Apply the formulas for the area of known planes to write the expression.
Detailed explanation
The area of a parallelogram is equal to the product of its base and height.
Therefore, the area of the parallelogram is \(5.6\) cm2.
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Solve problem 2 page 25 SBT Math 7 Creative horizon episode 2 – CTST
Write an algebraic expression representing the number of oranges arranged in the top 4 layers of the block in Figure 1.
Detailed instructions for solving Lesson 2
Solution method
We see the number of oranges in each layer arranged in a square. So we just need to count the number of oranges on the front of the cube and use the formula to calculate the area of the square.
Detailed explanation
We see the number of oranges in each layer arranged in a square.
Row 1 has the number of results \({1^2}\); row 2 has the result \({2^2}\); row 3 has the number of results \({3^2}\); row 4 has a result of \({4^2}\)
The algebraic expression for the number of oranges arranged in the top 4 layers of the block is \({1^2} + {2^2} + {3^2} + {4^2}\).
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Solution 3 page 25 SBT Math 7 Creative horizon episode 2 – CTST
Write an algebraic expression for the area of a rhombus whose first diagonal is 4 cm longer than the second.
Detailed instructions for solving Lesson 3
Solution method
Step 1: Set the hide and its condition.
Step 2: Represent other quantities according to the set implicits.
Step 3: Represent the relationship between the implicit quantities by mathematical operations.
Detailed explanation
Let the length of the first diagonal be \(x\,\left( {x > 0} \right)\) cm.
The length of the second diagonal is \(x – 4\) cm.
The algebraic expression for the area of the rhombus is \(\frac{1}{2}.x.\left( {x – 4} \right) = \frac{1}{2}{x^2 } – 2x\) cm2.
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Solution 4 page 25 SBT Math 7 Creative horizon episode 2 – CTST
Write an algebraic expression that represents the volume of a rectangular box whose length is 6 cm more than its width and 3 cm more than its height.
Detailed instructions for solving Lesson 4
Solution method
Step 1: Set the hide and its condition.
Step 2: Represent other quantities according to the set implicits.
Step 3: Represent the relationship between the implicit quantities by mathematical operations.
Detailed explanation
Let the length of the rectangle be \(x\,\left( {x > 0} \right)\) cm.
The width of the rectangle is \(x – 6\) cm
The height of the rectangle is \(x – 3\) cm.
The algebraic expression for the volume of a rectangular box is \(x\left( {x – 6} \right)\left( {x – 3} \right) = {x^3} – 3{x^ 2} – 6x + 18\) cm3.
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Solve problem 5 pages 25 SBT Math 7 Creative horizon episode 2 – CTST
Write an algebraic expression that represents:
a) Sum of \(3{x^2} + 1\) and \(5\left( {y – 2} \right)\);
b) Sum of the squares of \(a + 2\) and \(b – 1\).
Detailed instructions for solving Lesson 5
Solution method
Express algebraic expressions by mathematical operations required by the problem.
Detailed explanation
a) The sum of \(3{x^2} + 1\) and \(5\left( {y – 2} \right)\) is \(\left( {3{x^2} + 1} \) right) + 5\left( {y – 2} \right)\)
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b) The sum of the squares of \(a + 2\) and \(b – 1\) is \({\left( {a + 2} \right)^2} + {\left( {b – 1} \right)^2}\).
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Solution 6 page 25 SBT Math 7 Creative horizon episode 2 – CTST
The temperature in a town in the morning is x \(^\circ C\), at noon it increases by y \(^\circ C\), and in the evening it decreases by t \(^\circ C\)so with noon. Write an expression for the temperature at night. Calculate the temperature in the evening when \(x = 25;\,y = 5;\,t = 7\).
Detailed instructions for solving lesson 6
Solution method
Step 1: Represent the relationship between the quantities in the implicit by mathematical operations.
Step 2: Substitute the values of the variable to calculate the temperature in the evening.
Detailed explanation
The expression for the evening temperature is: \(x + y – t\).
The temperature at night when \(x = 25;\,y = 5;\,t = 7\) is \(25 + 5 – 7 = 23\)\(^\circ C\).
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Solution 7 page 25 SBT Math 7 Creative horizon episode 2 – CTST
Simplify the following expressions:
a) \(4\left( {2y – 3x} \right) – 3\left( {x – 2y} \right)\)
b) \({x^2} + 5y – 2y – 7{x^2}\)
Detailed instructions for solving lesson 7
Solution method
Use the rules of punctuation, breaking brackets learned to shorten expressions.
Detailed explanation
a) \(4\left( {2y – 3x} \right) – 3\left( {x – 2y} \right) = 8y – 12x – 3x + 6y = \left( {8y + 6y} \right) + \left( { – 12x – 3x} \right) = 14y – 15x\)
b) \({x^2} + 5y – 2y – 7{x^2} = \left( {{x^2} – 7{x^2}} \right) + \left( {5y – 2y} \right) = – 6{x^2} + 3y\)
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Solve problem 8 page 26 SBT Math 7 Creative horizon episode 2 – CTST
A rectangular garden has a width of a (m), length 6m more than width. The walkway is x (m) wide (see Figure 2). Write an expression that represents the area of the remainder of the garden. Calculate the area when a = 30 m, x = 1m.
Detailed instructions for solving Lesson 8
Solution method
Step 1: Set the hide and its condition.
Step 2: Represent other quantities according to the set implicits.
Step 3: Represent the relationship between the implicit quantities by mathematical operations.
Detailed explanation
Let the width of the rectangle be \(a\) (m) \(\left( {a > 0} \right)\)
According to the problem, we have: length is 6m more than width, so length of rectangle is \(a + 6\)(m)
The path area is \(far + x.\left( {a + 6} \right) – {x^2}\)
The area of the rectangle is \(a.\left( {a + 6} \right)\)
The remaining area is \(a.\left( {a – 6} \right) – \left[ {x.a + x.\left( {a + 6} \right) – {x^2}} \right]\).
Area left when \(a = 30\) m, \(x = 1\)m is \(a.\left( {a + 6} \right) – \left)[ {x.a + x.\left( {a + 6} \right) – {x^2}} \right] = 30.\left( {30 + 6} \right) – \left[ {1.30 + 1.\left( {30 + 6} \right) – {1^2}} \right] = 1015\) m2
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Solve lesson 9 page 26 SBT Math 7 Creative horizon episode 2 – CTST
Every morning you practice walking combined with running. The walking speed is 4 km/h and the running speed is 8 km/h.
a) Write an expression representing the distance that Thu walked x hours and ran y hours.
b) Calculate the distance when \(x = 15\) minutes and \(y = 30\) minutes.
Detailed instructions for solving Lesson 9
Solution method
Step 1: Express other quantities in terms of the given unknowns.
Step 2: Represent the relationship between the quantities in the implicit by mathematical operations.
Detailed explanation
a) The expression that represents the distance that Thu has walked x hours and run y is now \(4.x + 8.y\).
b) Convert \(x = 15\) minutes \( = 0.25\) hours and \(y = 30\) minutes \( = 0.5\)hours.
The distance that you travel is: \(4.0.25 + 8.0.5 = 5\)km.
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