Answer:
Step-by-step explanation:
The function f(x) is the another way of representing the y-values in the function
i.e
Then y-axis be labeled with f(x)
Also; the ordered pair is written as: (x, f(x)) instead of (x, y)
Given the equation:
replace y = f(x)
then;
therefore, the equation y=2x+3 using function notation is,
Answer:
a = 25m^2
b = 5m
d = 35.73 m^2
c = 7.94m
Step-by-step explanation:
First, remember that the area of a square of side length L is:
A = L^2
And for a triangle rectangle with catheti a and b, and hypotenuse H, we have the relation:
H^2 = a^2 + b^2 (Phytagorean's theorem)
Ok, let's look at the left image, we have a green triangle rectangle.
The bottom cathetus has a length equal to the side length of a square with area of 16m^2
Then the side length of that square (and the cathetus) is:
L^2 = 16m^2
L = √(16m^2) = 4m
The left cathetus has a length equal to the side length of a square of area = 9m^2
Then the side length of that cathetus is:
K^2 = 9m^2
K = √(9m^) = 3m
Then the catheti of the green triangle rectangle are:
4m and 3m
Then the hypotenuse of this triangle (b) is:
b^2 = (4m)^2 + (3m)^2
b^2 = 16m^2 + 9m^2 = 25m^2
b = √(25m^2) = 5m
And b is the side length of the red square, then the area of that square is:
a = b^2 = 25m^2
Now let's go to the other image.
Here we have an hypotenuse of side length H, such that:
H^2 = 144m^2
And we have a cathetus (the one adjacent to the green triangle) of side length L such that:
L^2 = 81m^2
The other cathetus will have a sidelength c, such that:
c^2 = area of the purple square
By the Pythagorean's theorem we have:
144m^2 = 81m^2 + c^2
144m^2 = 81m^2 + c^2
144m^2 - 81m^2 = c^2
63m^2 = c^2
(√63m^2) = c = 7.94m
And the area of a triangle rectangle is equal to the product between the catheti divided by two.
We know that one cathetus is equal to c = 7.94m
And the other on is equal to the square root of 81m^2
√(81m^2) = 9m
then the area of the triangle is:
d = (7.94m)*(9m)/2 = 35.73 m^2
Answer:
The simplified form is
Step-by-step explanation:
The goal of the exercise is to apply exponent properties. On this case we need to distribute the exponent to each expression of the fraction, that is to the numerator and to the denominator, using the distribution exponent property.
In a general way we can always distribute exponents over multiplication or division.
Simplifying using exponent distribution property.
Applying the property to the exercise give us
Lastly we know that so we get
And that is the simplified form of the given expression.