Rathan thinks all factors of even numbers are even. Which explains whether Rathan is correct? He is correct because the product of 2 and any number is even.
He is correct because the product of two even numbers is even.
He is incorrect because the product of an even number and an odd number is even.
He is incorrect because the product of two odd numbers is odd

Answers

Answer 1
Answer: The answer to this would be C, the third of the options. This is the only response that fully answers the question. For example, two factors of 42 would be 6 and 7, and both are not even, so given the absolute nature of Rathan's belief, there is no exception that can be made to make him correct. I hope this helps, and if it was sufficient to make you understand fully set it to the brainliest answer. Have a nice day! =)
Answer 2
Answer:

The correct answer is:

He is incorrect because the product of an even number and an odd number is even.  

Explanation:

When you multiply any number by an even number, whether it is even or odd, the product will be even.

This means that an even number can have an odd factor, and Rathan is incorrect.


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x<-4 is the solution to the inequality
x<-4 is the answer hope I helped

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Step-by-step explanation:

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To change .3 to a fraction you must put 3 over what denominator? A. 100
B. .1
C. 10
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Answers

its C 3/10 =.3

There u go
C.10 When there is 1 digit after the decimal it is over 10, 2 =digits over 100, etc. how ever many digits equals how many zeros

Who is SohCahToa Joe? on Csi geometry:trigonometry I need the answers.

Answers

SohCahToa is an acronym for the basic trigonometric functions which are sine, cosine, and tangent. Sine's value comes from the quotient of the opposite side and the hypotenuse. Cosine's value comes from the quotient of the adjacent side and the hypotenuse. Lastly, Tangent is the quotient of the opposite side and the adjacent side.

Point P has coordinates (2,3). Point Q is symmetric to point P with respect to the line x = 4. What are the coordinates of point Q. Im wondering how you solve this. I think it's (3,6) A. (2,5)
B. ( 5,3)
C. (3,6)
D. (6,3

Answers

You are given the first point (2, 3) and a line x = 4. If you go ahead and graph these, it makes it a little more clear when they say that point p is symmetric with point Q with respect to the line x = 4.

Line x = 4 is the line of symmetry, so point Q is reflected across it. Point p has an x value of 2 (2 away from the line of symmetry, so point Q's x value has to be 2 away on the other side... 6. The y value is unchanged so your point is

D. (6, 3)

How to find the perimeter and the area of a rectangle on a coordinate plane using the distance formula? : A(3,8) B(5,4) C(-4,-1) D(-6,3) Round to the nearest tenth if necessary.

Answers

I used some site to plot the points. 

First, let's recall the formulas for Perimeter and Area of a Rectangle.
Perimeter = 2(l+w)
Area = l×w
Also, the distance formula is
D = \sqrt{( x_(2)-x_(1))^2+{(y_(2)-y_(1))^2}

Now, we need to determine l and w.
So, the length is the distance of AD or BC
and the width is the distance of DC or AB
(I'll just use the sides that I've labelled for ease)

So first to determine the length, we need to calculate the distance of AD
Points are A(3,8) and D(-6,3) 
x_(1) =3, x_(2) =8, y_(1) =6, y_(2) =-3
D_(AD)= \sqrt{( 6-3)^2+{(-3-8)^2}
D_(AD)= √((3)^2+(-11)^2)
D_(AD)= √(9+121)
D_(AD)= √(130) ≈ 11.40
So the length of the rectangle is 11.40 units.

Now, the width!
So first to determine the width, we need to calculate the distance of DC
Points are D(-6,3) and C(-4,-1)
x_(1) =6, x_(2) =-4, y_(1) =-3, y_(2) =-1
D_(DC)= \sqrt{(-4-6)^2+{(-1- -3)^2}
<span>D_(DC)= \sqrt{(-4-6)^2+{(-1+3)^2}
D_(DC)= √((-10)^2+(2)^2)
D_(AD)= √(100+4)
D_(AD)= √(104) ≈ 10.20
So the width of the rectangle is ≈10.20 units.

Let's now solve for Perimeter and Area using
l = 11.40
w = 10.20

Perimeter = 2(l+w)
Perimter = 2(11.40+10.20)
Perimter = 2(21.60)
Perimter = 43.2 units

Area = l×w
Area = (11.40)(10.20)
Area = 116.28 
Area = 116.3 units² (rounding)

In conclusion, given points A(3,8) B(5,4) C(-4,-1) D(-6,3), the Perimeter is 43.2 units and the Area is  ≈116.3 units² using the distance formula.