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notsponge [240]
3 years ago
13

Which is a counterexample that shows the statement is false? If you square a number, then the answer is greater than the number.

Mathematics
1 answer:
Iteru [2.4K]3 years ago
7 0

Answer:

A. (\frac{1}{2})² = \frac{1}{2} × \frac{1}{2} = \frac{1}{4}

Step-by-step explanation:

Answer choice (A) is the counterexample because squaring number \frac{1}{2} results into number \frac{1}{4} which is less than \frac{1}{2} .

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Isabella decides to sell handmade stationery. She decides to sell 2 cards for $9. Which table below show the possible values of
kolbaska11 [484]

Answer:

B

Step-by-step explanation:

each time she sells 2 cards she gets $9 right?

so counting by 2's is the best route to go on and the only answer that displays that is answer B

5 0
3 years ago
Solve for the roots in simplest form using the quadratic formula:<br> 4x^2+5=−12x
MA_775_DIABLO [31]
1/5 or 5/2 either one works!!
6 0
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Help please will give brainliest
ser-zykov [4K]
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7 0
2 years ago
Sean operates a lawn-cutting business. He charges $15 for a single lawn and $25 for a front and back lawn. Sean made $475 from 2
topjm [15]

The system of equations that models the given situation is given as follows:

  • x + y = 25.
  • 15x + 25y = 475.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are given as follows:

  • Variable x: Number of single lawns cut.
  • Variable y: Number of front and back lawns cut.

He had 25 customers, hence:

x + y = 25.

He charges $15 for a single lawn and $25 for a front and back lawn, and made a total of $475, hence:

15x + 25y = 475

More can be learned about a system of equations at brainly.com/question/24342899

#SPJ1

3 0
2 years ago
We have 4different boxesand 6different objects. We want to distribute the objects into the boxes such that at no box is empty. I
Musya8 [376]

Answer:

Following are the solution to this question:

Step-by-step explanation:

They provide various boxes or various objects.  It also wants objects to be distributed into containers, so no container is empty.  All we select k objects of r to keep no boxes empty, which (r C k) could be done.  All such k artifacts can be placed in k containers, each of them in k! Forms. There will be remaining (r-k) objects. All can be put in any of k boxes.  Therefore, these (r-k) objects could in the k^{(r-k)} manner are organized.  Consequently, both possible ways to do this are

=\binom{r}{k} \times k! \times k^{r-k}\\\\=\frac{r! \times k^{r-k}}{(r-k)!}

Consequently, the number of ways that r objects in k different boxes can be arranged to make no book empty is every possible one

= \frac{r!k^{r-k}}{(r-k)!}

5 0
3 years ago
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