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2 years ago

Which of the following problems would NOT have a solution?

Mathematics

2 answers:

Vika [28.1K]2 years ago

6 0

The last one I think

Natasha2012 [34]2 years ago

6 0

Answer:

The correct option is D) Two pizzas are shared equally among zero people, and you want to know how much each person gets.

Step-by-step explanation:

Consider the provided information,

We need to identify the option which has no solution,

Consider the option A)

Six pizzas are shared equally among three people, and you want to know how much each person gets.

We can find how much each person gets by dividing the number or pizza with number of people:

Each person gets = 6/3 = 2

That means each person gets 2 pizza.

The problem has a solution.

Consider the option B)

Three pizzas are shared equally among two people, and you want to know how much each person gets.

We can find how much each person gets by dividing the number or pizza with number of people:

Each person gets = 3/2 = 1.5

That means each person gets 1.2 pizza.

The problem has a solution.

Consider the option C)

Zero pizzas are shared equally among three people, and you want to know how much each person gets.

We can find how much each person gets by dividing the number or pizza with number of people:

Each person gets = 0/3 = 0

That means each person gets 0 pizza.

The problem has a solution.

Consider the option D)

Two pizzas are shared equally among zero people, and you want to know how much each person gets.

We can find how much each person gets by dividing the number or pizza with number of people:

Each person gets = 2/0 = No solution

As we know any number divided by 0 has no solution.

Thus, the problem has no solution.

Hence, the correct option is D) Two pizzas are shared equally among zero people, and you want to know how much each person gets.

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3 years ago

Read 2 more answers

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

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3 years ago

Jacob buys 3 hot dogs and a $1 soda.he spent $5.50 in all.what is the cost of one hot dog?

Margaret [11]

It would be $1.50 because if the soda's $1 then subtract it and divide that by the quantity.

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3 years ago

Read 2 more answers

Please help me ASAP!

vodomira [7]

Answer:

A and B are parallel to D and C

A and D to B and C

basically do a line and expand them if they touch they are not parralelogram but if they don't they are

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2 years ago