HELP ME FOR NUMBER 16

A store had 112 diet sodas. The ratio of diet sodas to regular sodas was 8:4. How many regular sodas were there?

IceJOKER [234]

Answer:

56

Step-by-step explanation:

There are two ways the answer to this question can be determined

<u><em>Method 1 : the fast method </em></u>

We know that 8 is twice 4

4 x 2 = 8

The ratio of diet soda = 8

the ratio of regular sodas = 4

Diet sodas = 112

the number of regular sodas = 112 / 2 = 56

<u><em>Method 2 : The long method </em></u>

I would first determine the total number of diet and regular sodas. Let the total number be represented by d

from the question, the following equation can be derived :

(8/12) x d = 112

divide both sides of the equation by 12/8 to determine the value of d

d = 112 x (12/8) = 168

We can now derive a value for the number of regular soda

regular sodas = ( ratio of regular sodas / total soda) x total number of sodas

(4/12) x 168 = 56

3 0

3 years ago

(Geometry course question) Explain why it is not possible to construct an equilateral triangle that has

Nutka1998 [239]

Answer:

√3 is irrational

Step-by-step explanation:

The location of the third point of a triangle can be found using a rotation matrix to transform the coordinates of the given points.

<h3 /><h3>Location of point C</h3>

With reference to the attached figure, the slope of line AC is √3, an irrational number. This means the line AC <em>never passes through a point with integer coordinates</em>. (Any point with integer coordinates would be on a line with rational slope.)

<h3>Equilateral triangle</h3>

The line segments making up an equilateral triangle are separated by an angle of 60°. If two vertices are on grid squares, the third must be a rotation of one of them about the other through an angle of 60°. The rotation matrix is irrational, so the rotated point must have irrational coordinates.

The math of it is this. For rotation of (x, y) counterclockwise 60° about the origin, the transformation matrix is ...

$\left[\begin{array}{cc}\cos(60^\circ)&\sin(60^\circ)\\-\sin(60^\circ)&\cos(60^\circ)\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}x'\\y'\end{array}\right]$

Cos(60°) is rational, but sin(60°) is not. For any non-zero rational values of x and y, the sum ...

cos(60°)·x + sin(60°)·y

will be irrational.

As in the attached diagram, if one of the coordinates of the rotated point (B) is zero, then one of the coordinates of its image (C) will be rational. The other image point coordinate cannot be rational.