Volume ratio = 1331/729 which is the cube of the linear scale factor.
To find the linear scale factor, let find the cubic root of the numerator & the denominator:
∛1331 = ∛11³ = 11
& ∛729 = ∛9³ = 9
So the linear scale is 11/9 ==> then the ratio of their surface area will be:
11²/9² ==> 121/81.
Note, if you have a linear scale, then the surface will be the square othis scale & the volume will be the cube of the linear scale
The solution to the system of equations is (x, y) = (10, 14), where x is the number of 3-point questions and y is the number of 5-point questions. The appropriate choice is ...
The test contains 10 three-point questions and 14 five-point questions. (2nd selection)
Answer:
42.6cm
Step-by-step explanation:
The sum total of all the sides of the shape is the perimeter of the shape. Hence;
Perimeter = x+3 + 4 + x+9 + 2x+5
Perimeter = 4x + 21
Given
Area of the figure = 75cm^2
Area = (x+3)(2x+5) + (2x+1)(6)
75 = 2x²+5x+6x+15+12x+6
75 = 2x²+23x+21
2x²+23x+21-75 = 0
2x²+23x-54 = 0
x = -23±√23²-4(2)(-75)/4
x = -23±√529+600/4
x = -12±√1129/4
x = -12+33.6/4
x = 21.6/4
x = 5.4
Recall that;
Perimeter = 4x+21
Perimeter = 4(5.4) + 21
Perimeter = 21.6+21
Perimeter = 42.6cm
Hence the perimeter is 42.6cm
Answer: ( 0, 2 )
Step-by-step explanation:
You have to put coordinates of each graph on these equations y = -x + 2 and y = (1/2)x + 2. If putting coordinates satisfies both equations, then that coordinate will be the solution.
For example, let's put (0, 2) to equations.
y = -x + 2
2 = -0 + 2
2 = 2, true
y = (1/2)x + 2
2 = (1/2) × 0 + 2
2 = 2, true
So, ( 0, 2 ) is the solution.
Answer:

Step-by-step explanation:
Here we are given two coordinates through which our lines passes through. Now we are going to use the two point form to find the equation of the line.
The two point form is given as

Here we are given two coordinates . Thus replacing them in the formula and simplifying it will give us the equation of the line.




adding x and 6 on both sides we get

Hence this is our equation of the line passing through (5,6) & (4,7)