Start with parenthesis first.
so a+a+2a+48
Then add all like terms.
4a=48
Then solve for a by dividing 4 from 48.
a=12
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
Answer:
x = sqrt(105)
Step-by-step explanation:
Since this is a right triangle, we can use the Pythagorean theorem
a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse
16^2 + x^2 = 19^2
256 + x^2 = 361
x^2 = 361 -256
x^2 =105
Take the square root of each side
sqrt(x^2) = sqrt(105)
x = sqrt(105)
Area=height times base (for some prisms including cylinders)
vcylinder=hpir²
h=height
therefor pir²=base area
vcylinder=height times aeraofbase
given
h=9
v=324pi
324pi=9(basearea)
divide both sides by 9
36pi=areabase
the area of te base is 36pi square cm (put 36 in the blank since th pi is alredy there)
Answer:
20
Step-by-step explanation:
The lower quartile Q₁ is the value at the left side of the box, that is
Q₁ = 20