Seahorses change color and swim together in predetermined ways before mating. These behaviors are called
2 answers:
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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.
You can find the simplest form by dividing both the numerator and denominator by the GCF (greatest common factor). 100 divides into 100 and 1,000 equally so that can be the GCF used to reduce the fraction into its simplest form.
The simplest form is .