All categories

3 years ago

There are 24 counters. How many counters could you add without having to regroup? Explain.

1 answer:

3 0

The answer to this question would have to be 5. In the statement given we can add 5 more that is 24 _ 5 equal to 29. Once the counter reaches 30, one must have to regroup by exchanging 10 ones for a 10 bar.Hence the answer to this problem would have to be 5 counters

You might be interested in

HELP I DON'T UNDERSTAND THIS!!!!!

vladimir2022 [97]

316

61+17+17+17+17+17+17+17+17+17+17+17+17+17+17+17

5 0

4 years ago

Read 2 more answers

20 POINTS

RUDIKE [14]

<em>The correct expressions are as follows:</em>

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $343$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49^{\frac{2}{10}} \cdot 7^{\frac{1}{5}}$

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall following formula about Exponents and Surds:

$\boxed { \sqrt { x } = x ^ { \frac{1}{2} } }$

$\boxed { (a ^ b) ^ c = a ^ { b . c } }$

$\boxed {a ^ b \div a ^ c = a ^ { b - c } }$

$\boxed {\log a + \log b = \log (a \times b) }$

$\boxed {\log a - \log b = \log (a \div b) }$

<em>Let us tackle the problem!</em>

$\texttt{ }$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5}} \cdot (7^2)^{\frac{7}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5}} \cdot (7)^{2\times \frac{7}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = \boxed{7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{1}{5} + \frac{14}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{\frac{15}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = 7^{3}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}} = \boxed{343}$

$\texttt{ }$

<em>From the results above, it can be concluded that the correct statements are:</em>

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $343$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Equivalent $7^{\frac{1}{5}} \cdot 7^{\frac{14}{5}}$

$7^{\frac{1}{5}} \cdot 49^{\frac{7}{5}}$ Not Equivalent $49^{\frac{2}{10}} \cdot 7^{\frac{1}{5}}$

$\texttt{ }$

<h3>Learn more</h3>

Coefficient of A Square Root : brainly.com/question/11337634
The Order of Operations : brainly.com/question/10821615
Write 100,000 Using Exponents : brainly.com/question/2032116

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Exponents and Surds

Keywords: Power , Multiplication , Division , Exponent , Surd , Negative , Postive , Value , Equivalent , Perfect , Square , Factor.

4 0

4 years ago

Read 2 more answers

Is 8.26 an irrational number

densk [106]

Answer:

no it is not

Step-by-step explanation:

6 0

3 years ago

Someone please help!!

Wittaler [7]

A. (X+7)^2
B.(x-8)^2..............

3 0

3 years ago

Two 6-sided dice are rolled. What is the probability that the sum of the two numbers of the dice will be less than 6?

Elena-2011 [213]

Answer:

I believe it would just be 5/6.

Step-by-step explanation:

In a six-sided dice, there are five numbers less than 6!

6 0

3 years ago

Read 2 more answers