Q = 16
View the attached image for work!
Answer:
The width is 50 yards and the length is 141 yards.
Step-by-step explanation:
Let's call: L the length of the field and W the width of the field.
From the sentence, the perimeter of the rectangular playing field is 382 yards we can formulate the following equation:
2L + 2W = 382
Because the perimeter of a rectangle is the sum of two times the length with two times the width.
Then, from the sentence, the length of the field is 9 yards less than triple the width, we can formulate the following equation:
L = 3W - 9
So, replacing this last equation on the first one and solving for W, we get:
2L + 2W = 382
2(3W - 9) + 2W = 382
6W -18 +2W = 382
8W - 18 = 382
8W = 382 + 18
8W = 400
W = 400/8
W = 50
Replacing W by 50 on the following equation, we get:
L = 3W - 9
L = 3(50) - 9
L = 141
So, the width of the rectangular field is 50 yards and the length is 141 yards.
Answer:
47.6m.
Step by step solution:
Perimeter of a triangle = base + 2 . length____(1)
Area of a triangle = 1/2 . base . diagonal
108 = 1/2 . base . 15
multiplying both sides by 2:
216 = 15 . base
dividing both sides by 15:
base = 14.4m
But the diagonal divides the triangle into two
right angle triangles each with the same length (hypotenuse),base and diagonal(height).
Taking one right angle triangle:
And using pythagoras theorem;
length² = base² + diagonal ²
length² = 7.2² + 15²
Note: Base of each right angle triangle is 7.2 which would sum up to be 14.4 the base of the full triangle.
length² = 276.84
taking the square root of both sides:
length = 16.6m
Putting the values of the base and length into equation (1).
Perimeter of the triangle = 14.4 + 2 . 16.6
Note: We are dealing with the whole triangle
now hence the base is 14.4m.
Perimeter of the triangle = 14.4 + 33.2 = 47.6m.
Step-by-step explanation:
f(x) = x² + x + 3/4
in general, such a quadratic function is defined as
f(x) = a×x² + b×x + c
the solution for finding the values of x where a quadratic function value is 0 (there are as many solutions as the highest exponent of x, so 2 here in our case)
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 1
c = 3/4
x = (-1 ± sqrt(1² - 4×1×3/4))/(2×1) =
= (-1 ± sqrt(1 - 3))/2 = (-1 ± sqrt(-2))/2 =
= (-1 ± sqrt(2)i)/2
x1 = (-1 + sqrt(2)i) / 2
x2 = (-1 - sqrt(2)i) / 2
remember, i = sqrt(-1)
f(x) has no 0 results for x = real numbers.
for the solution we need to use imaginary numbers.
