Question

# Show the conjecture is false by finding a counterexample.

If the product of two numbers is even, then the two numbers must be even.

## The correct answer is: the given conjecture is false

### We have given a statement

If the product of two numbers is even, then the two numbers must be even.

We have to show the given conjecture is false.

Let us take two even numbers 6 and 8

Product of considered numbers = 6 x 8 = 48

Product we get is an even number , so the given statement is true for considered values.

If we consider one odd and one even number , 5 and 6

Product = 5 x 6 = 30

So, the product obtained is an even which contradicts the given statement

So, We can say that if the product of two numbers is even, then the two numbers need be even.

So, the given conjecture is false.

### Related Questions to study

### Show the conjecture is false by finding a counterexample.

If a + b = 0, then a = b = 0.

We have given a statement

If a + b = 0, then a = b = 0

We have to show the given conjecture is false.

Let us take a = 0 and b = 0

Then a + b = 0 + 0 = 0

Given statement is true for considered values of a and b.

Let us take a = 3 and b = - 3

Then a + b = 3 + (- 3) = 3 – 3 = 0

From this example we can say that it is not necessary that and b should be 0 to get sum of a and b to be 0 .

Therefore, the given conjecture is false.

### Show the conjecture is false by finding a counterexample.

If a + b = 0, then a = b = 0.

We have given a statement

If a + b = 0, then a = b = 0

We have to show the given conjecture is false.

Let us take a = 0 and b = 0

Then a + b = 0 + 0 = 0

Given statement is true for considered values of a and b.

Let us take a = 3 and b = - 3

Then a + b = 3 + (- 3) = 3 – 3 = 0

From this example we can say that it is not necessary that and b should be 0 to get sum of a and b to be 0 .

Therefore, the given conjecture is false.

### calculate the surface area of a cube and volume of a cube with the given diagonal length of 15cm

D= 15 cm

We have to find the Surface area and volume of the cube.

We know that Diagonal of the cube = a

Where a is side of the cube

Comparing with the given value we get

a = 15 cm

Therefore , Surface area of cube = 6a^{2}

= 6(15)^{2}

= 6(225)

= 1350

And, volume of the given cube = a^{3}

= (15)^{3}

= 3375

Therefore the correct option is c)1350, 3375.

### calculate the surface area of a cube and volume of a cube with the given diagonal length of 15cm

D= 15 cm

We have to find the Surface area and volume of the cube.

We know that Diagonal of the cube = a

Where a is side of the cube

Comparing with the given value we get

a = 15 cm

Therefore , Surface area of cube = 6a^{2}

= 6(15)^{2}

= 6(225)

= 1350

And, volume of the given cube = a^{3}

= (15)^{3}

= 3375

Therefore the correct option is c)1350, 3375.

### Show the conjecture is false by finding a counterexample.

If the product of two numbers is positive, then the two numbers must both be positive.

Solution:-

We have given a statement If the product of two numbers is positive, then the two numbers must both be positive.

If we consider two positive numbers 7 and 9

Then , their product, 7 x 9 = 63

Hence, the statement is correct for positive integers

Suppose we take two negative integers - 7 and - 9

Then, their product, - 7 x - 9 = 63

So,The product of any two positive numbers is positive, and the product of any two negative numbers is also positive.

Therefore, the given statement is false.

### Show the conjecture is false by finding a counterexample.

If the product of two numbers is positive, then the two numbers must both be positive.

Solution:-

We have given a statement If the product of two numbers is positive, then the two numbers must both be positive.

If we consider two positive numbers 7 and 9

Then , their product, 7 x 9 = 63

Hence, the statement is correct for positive integers

Suppose we take two negative integers - 7 and - 9

Then, their product, - 7 x - 9 = 63

So,The product of any two positive numbers is positive, and the product of any two negative numbers is also positive.

Therefore, the given statement is false.

### Show the conjecture is false by finding a counterexample.

The square root of any positive integer x is always less than x.

The square root of any positive integer x is always less than

Let us take any positive integer x

We consider that the square root of any interger x is always less than x

But if we consider integer 1 then,

The square root of 1 is 1.

Which is not less than 1

So, the given statement is false

### Show the conjecture is false by finding a counterexample.

The square root of any positive integer x is always less than x.

The square root of any positive integer x is always less than

Let us take any positive integer x

We consider that the square root of any interger x is always less than x

But if we consider integer 1 then,

The square root of 1 is 1.

Which is not less than 1

So, the given statement is false

### Show the conjecture is false by finding a counterexample.

All prime numbers are odd.

All prime numbers are odd

Before deciding we will know the definition of prime numbers

Definition - a prime number has only 2 factors - itself and 1.

Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.

Therefore, it shows that All prime numbers are not odd.

### Show the conjecture is false by finding a counterexample.

All prime numbers are odd.

All prime numbers are odd

Before deciding we will know the definition of prime numbers

Definition - a prime number has only 2 factors - itself and 1.

Hence the smallest natural prime number is 2, and the only on that is even. All other prime numbers are odd, and there are infinitely many prime numbers.

Therefore, it shows that All prime numbers are not odd.

### Complete the conjecture.

The sum of the first n odd positive integers is ____.

a

_{n}= a + (n – 1) × d

To find the sum of arithmetic progression, we have to know the first term, the number of terms and the common difference between each term. Then use the formula given below:

S = n/2[2a + (n − 1) × d]

To find: Sum of first n odd natural numbers

The first n odd natural numbers are given by 1,3,5,7,9…… (2n – 1) and this forms an AP.

- a = 1
- d = 2
- t
_{n}= (2n – 1)

S = n/2[2a + (n − 1) × d]

S = n /2 [2 + 2n – 2]

S = n /2 [2n]

S = n

^{2}

The sum of first n odd natural numbers is n

^{2}.

### Complete the conjecture.

The sum of the first n odd positive integers is ____.

a

_{n}= a + (n – 1) × d

To find the sum of arithmetic progression, we have to know the first term, the number of terms and the common difference between each term. Then use the formula given below:

S = n/2[2a + (n − 1) × d]

To find: Sum of first n odd natural numbers

The first n odd natural numbers are given by 1,3,5,7,9…… (2n – 1) and this forms an AP.

- a = 1
- d = 2
- t
_{n}= (2n – 1)

S = n/2[2a + (n − 1) × d]

S = n /2 [2 + 2n – 2]

S = n /2 [2n]

S = n

^{2}

The sum of first n odd natural numbers is n

^{2}.

### A square is drawn with the length of the side equal to the diagonal of the cube. if the square area is 72075 cm^{2}, then find the side of the cube?

Area of square=side × side=a²

Diagonal of cube= unit

Solution:- The length of the square and the diagonal of the cube are coincided. Therefore, the value of the two will be equal.

- We are given that,

Side of square=Diagonal of cube

*Let the edge of the cube be 'a' cm.*

As we know,

length of the diagonal of cube= ×edge of the cube=a cm

- The length of the side of the square is equal to the length of the diagonal of the cube.

- Area of square(drawn on diagonal of cube)

- We know that, area of square drawn on the diagonal of a cube= 72,075 cm²

Therefore , the side of the cube is 155 cm.

### A square is drawn with the length of the side equal to the diagonal of the cube. if the square area is 72075 cm^{2}, then find the side of the cube?

Area of square=side × side=a²

Diagonal of cube= unit

Solution:- The length of the square and the diagonal of the cube are coincided. Therefore, the value of the two will be equal.

- We are given that,

Side of square=Diagonal of cube

*Let the edge of the cube be 'a' cm.*

As we know,

length of the diagonal of cube= ×edge of the cube=a cm

- The length of the side of the square is equal to the length of the diagonal of the cube.

- Area of square(drawn on diagonal of cube)

- We know that, area of square drawn on the diagonal of a cube= 72,075 cm²

Therefore , the side of the cube is 155 cm.

### What are all the possible values of b for which is it factorable using only integer coefficients and constant?

Given, is it factorable so be must be sum of factors of (-8)(3)

Factors of -24 are

(1,-24) b can be -24 + 1 = -23

(2,-12) b can be -12 + 2 = -10

(3,-8) b can be -8 + 3 = -5

(4,-6) b can be 4 - 6 = -2

(6,-4) b can be 6 - 4 = 2

(8,-3) b can be 8 - 3 = 5

(12,-2) b can be 12 - 2 = 10

(24,-1) b can be 24 - 1 = 23

So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}

### What are all the possible values of b for which is it factorable using only integer coefficients and constant?

Given, is it factorable so be must be sum of factors of (-8)(3)

Factors of -24 are

(1,-24) b can be -24 + 1 = -23

(2,-12) b can be -12 + 2 = -10

(3,-8) b can be -8 + 3 = -5

(4,-6) b can be 4 - 6 = -2

(6,-4) b can be 6 - 4 = 2

(8,-3) b can be 8 - 3 = 5

(12,-2) b can be 12 - 2 = 10

(24,-1) b can be 24 - 1 = 23

So, there are 8 possible values of b they are {-23, -10, -5, -2, 2, 5, 10, 23}

### Find a counterexample to show that the following conjecture is false.

Conjecture: The square of any integer is always greater than the integer.

We have to prove that given statement false.

Number obtained when a number is multiplied by itself three times is called a cube number.

If m = n², then m is a perfect square where m and n are natural numbers.

Example: consider 1

Square of 1 = 1

Example: consider 2

Square of 2 = 4

Therefore, the square of every natural number is not always greater than the number itself.

Therefore, given statement is false.

### Find a counterexample to show that the following conjecture is false.

Conjecture: The square of any integer is always greater than the integer.

We have to prove that given statement false.

Number obtained when a number is multiplied by itself three times is called a cube number.

If m = n², then m is a perfect square where m and n are natural numbers.

Example: consider 1

Square of 1 = 1

Example: consider 2

Square of 2 = 4

Therefore, the square of every natural number is not always greater than the number itself.

Therefore, given statement is false.

### Three blocks that are in the shape of a cube with each side 3.2 cm are attached end to end. Calculate the T.S.A of the resulting cuboid?

When three cubes of side 3.2 cm are joined, the resulting shape is a cuboid of

length l = 3.2×3=9.6 cm;

breadth b = 3.2 cm

and height h= 3.2 cm

So, total surface area of a cuboid = 2(lb + bh + lh)

=2(9.6×3.2+3.2×3.2+9.6×3.2)

=143.36 cm

^{2}

Therefore the total surface area of the cuboid obtained is 143.36 cm

^{2}

Therefore the correct option is a)143.36

### Three blocks that are in the shape of a cube with each side 3.2 cm are attached end to end. Calculate the T.S.A of the resulting cuboid?

When three cubes of side 3.2 cm are joined, the resulting shape is a cuboid of

length l = 3.2×3=9.6 cm;

breadth b = 3.2 cm

and height h= 3.2 cm

So, total surface area of a cuboid = 2(lb + bh + lh)

=2(9.6×3.2+3.2×3.2+9.6×3.2)

=143.36 cm

^{2}

Therefore the total surface area of the cuboid obtained is 143.36 cm

^{2}

Therefore the correct option is a)143.36

### The area of a playground is . Without removing common factors, factor to possible dimensions of the playground. How are the side-lengths related? What value would you need to subtract from the longer side and add to the shorter side for the playground to be a square?

Ans :- 6x + 4y and 6x - 4y are the dimensions of a rectangular playground . 4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square .

Explanation :-

Given,

Write

As

Here

We get

As the garden is rectangular , we can write area A = length × breadth

The possibility is length = 6x + 4y and breadth = 6x - 4y (or) viceversa

4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square

### The area of a playground is . Without removing common factors, factor to possible dimensions of the playground. How are the side-lengths related? What value would you need to subtract from the longer side and add to the shorter side for the playground to be a square?

Ans :- 6x + 4y and 6x - 4y are the dimensions of a rectangular playground . 4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square .

Explanation :-

Given,

Write

As

Here

We get

As the garden is rectangular , we can write area A = length × breadth

The possibility is length = 6x + 4y and breadth = 6x - 4y (or) viceversa

4y needs to be subtracted from the longer side(6x + 4y) and adding it to the shorter side(6x - 4y) gives the playground to be a square

### What is the first three numbers in the pattern?

−, −, −, 64, 128, 256, …

a

_{n}= a + (n – 1) × d

Solution:-

We have given a pattern

−, −, −, 64, 128, 256, …

We have to find the first 3 terms of the given pattern

Fourth term is 64 and fifth is 128 and sixth is 256

On analysis we get that , the next term is double of the previous term

Therefore the first three terms will be

Third term = 64 / 2 = 32

Second term = 32 / 2 = 16

First term = 16/2 = 8

Therefore, the pattern will be 8 , 16 , 32 , 64, 128 , 256 ,…

### What is the first three numbers in the pattern?

−, −, −, 64, 128, 256, …

a

_{n}= a + (n – 1) × d

Solution:-

We have given a pattern

−, −, −, 64, 128, 256, …

We have to find the first 3 terms of the given pattern

Fourth term is 64 and fifth is 128 and sixth is 256

On analysis we get that , the next term is double of the previous term

Therefore the first three terms will be

Third term = 64 / 2 = 32

Second term = 32 / 2 = 16

First term = 16/2 = 8

Therefore, the pattern will be 8 , 16 , 32 , 64, 128 , 256 ,…

### Calculate the maximum number of chocolates of size 2cmx3 cmx5 cm that can be kept in a rectangular box of dimensions 6cmx3cmx15 cm

Size of chocolate = 2cm 3cm 5cm

Size of box = 6cm 3cm 15cm

We have to find out the maximum number of chocolates that can be kept in the given box,

So, we can get our required answer by dividing Volume of box by volume of one soap.

Maximum number of chocolates = Volume of box / volume of one soap

Therefore, the correct option is a) 9.

### Calculate the maximum number of chocolates of size 2cmx3 cmx5 cm that can be kept in a rectangular box of dimensions 6cmx3cmx15 cm

Size of chocolate = 2cm 3cm 5cm

Size of box = 6cm 3cm 15cm

We have to find out the maximum number of chocolates that can be kept in the given box,

So, we can get our required answer by dividing Volume of box by volume of one soap.

Maximum number of chocolates = Volume of box / volume of one soap

Therefore, the correct option is a) 9.

### Make and test a conjecture about the sign of the cube of negative integers

The cube of a number is the number multiplied by itself thrice. For the negative numbers, cubes are negative since the negative numbers are multiplied thrice. When two negative numbers are squared, the result is positive. However, when the negative numbers are cubed, the result is negative

(- 3)

^{3}

= (-3) × (- 3) = 9

= 9 × (- 3)

= - 27

Hence, the cube of (-3) is -27.

Therefore, the sign of cube of negative intergers is negative.

### Make and test a conjecture about the sign of the cube of negative integers

The cube of a number is the number multiplied by itself thrice. For the negative numbers, cubes are negative since the negative numbers are multiplied thrice. When two negative numbers are squared, the result is positive. However, when the negative numbers are cubed, the result is negative

(- 3)

^{3}

= (-3) × (- 3) = 9

= 9 × (- 3)

= - 27

Hence, the cube of (-3) is -27.

Therefore, the sign of cube of negative intergers is negative.

### Given the area of a square, factor it to find the side length.

^{2}. find the side of the square by factoring the area .

Ans:- 12x-1 is the side length of the square.

Explanation:-

Given , Area A =

Write

we get A =

Applying

We get A =

But, area of square A =

∴ length of side is 12x - 1 .

### Given the area of a square, factor it to find the side length.

^{2}. find the side of the square by factoring the area .

Ans:- 12x-1 is the side length of the square.

Explanation:-

Given , Area A =

Write

we get A =

Applying

We get A =

But, area of square A =

∴ length of side is 12x - 1 .