In order to prove Rathan wrong, we only need one counterexample. Take the number 6. 6 is even, but it has the odd number 3 as a factor, so clearly, not all factors of even numbers are even.

7 0

3 years ago

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The product of x and 7

EastWind [94]

Write that as 7x. Writing it as x7 could lead to confusion. 7*x would be OK.

8 0

3 years ago

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During a football game a concession stand sold a family three hamburgers and two hotdogs for a total of $13 it's sold another fa

guajiro [1.7K]

Answer:

The price of each hamburger is $3

The price of each hot dogs is $2 .

Step-by-step explanation:

Given as :

The total price of 3 hamburger and 2 hot dogs = $13

The total price of 2 hamburger and 5 hot dogs = $16

Let The price of each hamburger = $x

Let The price of each hot dogs = $y

Now, According to question

3 x + 2 y = 13 .........A

2 x + 5 y = 16 .......B

Now, Solving to eq A and B

3 × (2 x + 5 y ) - 2 × (3 x + 2 y ) = 3 × 16 - 2 × 13

Or, (6 x + 15 y) - (6 x + 4 y) = 48 - 26

Or, (6 x - 6 x) + (15 y - 4 y) = 22

Or, 0 + 11 y = 22

∴ y = $\dfrac{22}{11}$

i.e y = $2

so, The price of each hot dogs = y = $2

Now, Put the value of y into eq B

i.e 2 x + 5 y = 16

or, 2 x + 5 × 2 = 16

or, 2 x = 16 - 10

or, 2 x = 6

∴ x = $\dfrac{6}{2}$

i.e x = $3

So, The price of each hamburger = x = $3

Hence, The price of each hamburger is $3 and The price of each hot dogs is $2 . Answer

3 0

3 years ago