Solve for x to the nearest tenth
2 answers:
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After plotting the quadrilateral in a Cartesian plane, you can see that it is not a particular quadrilateral. Hence, you need to divide it into two triangles. Let's take ABC and ADC.
The area of a triangle with vertices known is given by the matrix
M =![\left[\begin{array}{ccc} x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20x_%7B1%7D%26y_%7B1%7D%261%5C%5Cx_%7B2%7D%26y_%7B2%7D%261%5C%5Cx_%7B3%7D%26y_%7B3%7D%261%5Cend%7Barray%7D%5Cright%5D%20)
Area = 1/2· | det(M) |
= 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
= 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |
Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
= 1/2· | -5·(1) - 4·(-13) - 1·(12) |
= 1/2 | 35 |
= 35/2
Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
= 1/2· | -5·(11) + 4·(-13) - 1·(2) |
= 1/2 | -109 |
<span> = 109/2</span>
The total area of the quadrilateral will be the sum of the areas of the two triangles:
A(ABCD) = A(ABC) + A(ADC)
= 35/2 + 109/2
= 72
The area of a triangle with vertices known is given by the matrix
M =
Area = 1/2· | det(M) |
= 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
= 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |
Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
= 1/2· | -5·(1) - 4·(-13) - 1·(12) |
= 1/2 | 35 |
= 35/2
Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
= 1/2· | -5·(11) + 4·(-13) - 1·(2) |
= 1/2 | -109 |
<span> = 109/2</span>
The total area of the quadrilateral will be the sum of the areas of the two triangles:
A(ABCD) = A(ABC) + A(ADC)
= 35/2 + 109/2
= 72
Answer:
a =
Step-by-step explanation:
Given:
f(x) = log(x)
and,
f(kaa) = kf(a)
now applying the given function, we get
⇒ log(kaa) = k × log(a)
or
⇒ log(ka²) = k × log(a)
Now, we know the property of the log function that
log(AB) = log(A) + log(B)
and,
log(Aᵇ) = b × log(A)
Thus,
⇒ log(k) + log(a²) = k × log(a) (using log(AB) = log(A) + log(B) )
or
⇒ log(k) + 2log(a) = k × log(a) (using log(Aᵇ) = b × log(A) )
or
⇒ k × log(a) - 2log(a) = log(k)
or
⇒ log(a) × (k - 2) = log(k)
or
⇒ log(a) = (k - 2)⁻¹ × log(k)
or
⇒ log(a) = (using log(Aᵇ) = b × log(A) )
taking anti-log both sides
⇒ a =