If an arithmetic sequence has a 10th term of 17 and a 14th term of 30. find the common difference (d).
1 answer:
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Answers:
1. 8 m
2. 17 m
3. 7 cm
4. 2 s
Explanations:
1. Let x = length of the base
x + 6 = height of the base
Then, the area of the triangle is given by
(Area) = (1/2)(base)(height)
56 = (1/2)(x)(x + 6)
56 = (1/2)(x² + 6x)
Using the symmetric property of equations, we can interchange both sides of equations so that
(1/2)(x² + 6x) = 56
Multiplying both sides by 2, we have
x² + 6x = 112
The right side should be 0. So, by subtracting both sides by 112, we have
x² + 6x - 112 = 112 - 112
x² + 6x - 112 = 0
By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation becomes
(x - 8)(x +14) = 0
So, either
x - 8 = 0 or x + 14 = 0
Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.
2. Let x = length of increase in both length and width of the rectangular garden
Then,
14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden
So,
(Area of the new garden) = (length of the new garden)(width of the new garden)
255 = (14 + x)(12 + x) (1)
Note that
(14 + x)(12 + x) = (x + 14)(x + 12)
= x(x + 14) + 12(x + 14)
= x² + 14x + 12x + 168
= x² + 26x + 168
So, the equation (1) becomes
255 = x² + 26x + 168
By symmetric property of equations, we can interchange the side of the previous equation so that
x² + 26x + 168 = 255
To make the right side becomes 0, we subtract both sides by 255:
x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0
To solve the preceding equation, we use the quadratic formula.
First, we let
a = numerical coefficient of x² = 1
Note: if the numerical coefficient is hidden, it is automatically = 1.
b = numerical coefficient of x = 26
c = constant term = - 87
Then, using the quadratic formula
So,
Since x represents the amount of increase, x should be positive.
Hence x = 3.
Therefore, the length of the new garden is given by
14 + x = 14 + 3 = 17 m.
3. The area of the shaded region is given by
(Area of shaded region) = π(outer radius)² - π(inner radius)²
= π(2x)² - π6²
= π(4x² - 36)
Since the area of the shaded region is 160π square centimeters,
π(4x² - 36) = 160π
Dividing both sides by π, we have
4x² - 36 = 160
Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.
Adding both sides of the equation by 36, we have
4x² - 36 + 36 = 160 + 36
4x² = 196
Then, we divide both sides by 4 so that
x² = 49
Taking the square root of both sides, we have
Note: If we take the square root of both sides, we need to add the plus minus sign
because equations involving x² always have 2 solutions.
So, x = 7 or x = -7.
But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.
Hence x = 7 cm
4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,
h(t) = 0 (1)
Moreover, since
and 
,
h(t) = -16t² + 27t + 10 (2)
Since the right side of the equations (1) and (2) are both equal to h(t), we can have
-16t² + 27t + 10 = 0
To solve this equation, we'll use the quadratic formula.
Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula.
First, we let
a = numerical coefficient of t² = -16
b = numerical coefficient of t = 27
c = constant term = 10
Then, using the quadratic formula
So,
Since t represents the amount of time, t should be positive.
Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.
1. 8 m
2. 17 m
3. 7 cm
4. 2 s
Explanations:
1. Let x = length of the base
x + 6 = height of the base
Then, the area of the triangle is given by
(Area) = (1/2)(base)(height)
56 = (1/2)(x)(x + 6)
56 = (1/2)(x² + 6x)
Using the symmetric property of equations, we can interchange both sides of equations so that
(1/2)(x² + 6x) = 56
Multiplying both sides by 2, we have
x² + 6x = 112
The right side should be 0. So, by subtracting both sides by 112, we have
x² + 6x - 112 = 112 - 112
x² + 6x - 112 = 0
By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation becomes
(x - 8)(x +14) = 0
So, either
x - 8 = 0 or x + 14 = 0
Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.
2. Let x = length of increase in both length and width of the rectangular garden
Then,
14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden
So,
(Area of the new garden) = (length of the new garden)(width of the new garden)
255 = (14 + x)(12 + x) (1)
Note that
(14 + x)(12 + x) = (x + 14)(x + 12)
= x(x + 14) + 12(x + 14)
= x² + 14x + 12x + 168
= x² + 26x + 168
So, the equation (1) becomes
255 = x² + 26x + 168
By symmetric property of equations, we can interchange the side of the previous equation so that
x² + 26x + 168 = 255
To make the right side becomes 0, we subtract both sides by 255:
x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0
To solve the preceding equation, we use the quadratic formula.
First, we let
a = numerical coefficient of x² = 1
Note: if the numerical coefficient is hidden, it is automatically = 1.
b = numerical coefficient of x = 26
c = constant term = - 87
Then, using the quadratic formula
So,
Since x represents the amount of increase, x should be positive.
Hence x = 3.
Therefore, the length of the new garden is given by
14 + x = 14 + 3 = 17 m.
3. The area of the shaded region is given by
(Area of shaded region) = π(outer radius)² - π(inner radius)²
= π(2x)² - π6²
= π(4x² - 36)
Since the area of the shaded region is 160π square centimeters,
π(4x² - 36) = 160π
Dividing both sides by π, we have
4x² - 36 = 160
Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.
Adding both sides of the equation by 36, we have
4x² - 36 + 36 = 160 + 36
4x² = 196
Then, we divide both sides by 4 so that
x² = 49
Taking the square root of both sides, we have
Note: If we take the square root of both sides, we need to add the plus minus sign
So, x = 7 or x = -7.
But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.
Hence x = 7 cm
4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,
h(t) = 0 (1)
Moreover, since
h(t) = -16t² + 27t + 10 (2)
Since the right side of the equations (1) and (2) are both equal to h(t), we can have
-16t² + 27t + 10 = 0
To solve this equation, we'll use the quadratic formula.
Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula.
First, we let
a = numerical coefficient of t² = -16
b = numerical coefficient of t = 27
c = constant term = 10
Then, using the quadratic formula
So,
Since t represents the amount of time, t should be positive.
Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.