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2 years ago

Can I get some help please?

Mathematics

1 answer:

mamaluj [8]2 years ago

3 0

Answer:

1) 68

/ \

4 17

2 2

68=2*2*17

2) 104

/ \

4 26

/\ /\

2 2 2 13

104=2*2*2*13

3) 78

2 39

3 13

78=2*3*13

4) 225

9 25

/\ /\

3 3 5 5

225=3*3*5*5

Second Part:

1) 45

3 15

3 5

45=3*3*5

2) 70

2 35

5 7

70=2*5*7

3) 54

9 6

/\ /\

3 3 2 3

54=3*3*2*3

4) 126

3 42

3 14

7 2

126=3*3*7*2

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125. Albert surveyed a class of 25 students on sports. 5 kids love baseball. 7 kids love basketball. 10 kids

MAXImum [283]

Answer:

3 students

Step-by-step explanation:

since the<em> </em>total number of students is 25,when you add those that like baseball, basketball and football the total number must be 25 but in this case it's 22 meaning 2 student liked neither.

7+5+10+x=25

x=25-22

I hope this helps

4 0

3 years ago

What is 11/9 reduced to a mixed number?? i cant figure this out

vodomira [7]

Answer:

11/9

=9/9+2/9

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3 0

3 years ago

Read 2 more answers

(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.