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pickupchik [31]
3 years ago
8

Jodi liked to collect stamps. On 3 different days, she bought 6 stamps. Then she decided to sell 4 of them. After that she bough

t 2 sets of 5 stamps. She figured out how many she had total, and then she divided that total into 3 groups to put into stamp books. How many did she put in each book?
Mathematics
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

4

Step-by-step explanation:

6-4=2

2x5=10

10+2=12

12 divided by 3=4

She put 4 in each book.

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Agata [3.3K]
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Use the Pythagorean Theorem to find the length of the missing side of the right triangle. Then find the value of each of the six
olganol [36]

Answer:

The length of the missing side can be calculated by the following steps;

Step-by-step explanation:

5 0
3 years ago
Find the value of c that makes the expression a perfect square trinomial.<br><br> x2 + 4x + c
masha68 [24]

the value of c that makes the expression a perfect square binomial is c=4 .

<u>Step-by-step explanation:</u>

Here we have , an expression x2 + 4x + c or , x^2 + 4x + c . We need to find  the value of c that makes the expression a perfect square binomial. Let's find out:

We have ,  x^2 + 4x + c

⇒ x^2+4x+c

⇒ x^2+2(2)x+c

Now , we know that (a+b)^2=a^2+2(a)(b)+b^2

Comparing above equation  , to  x^2+2(2)x+c we get ;

⇒  x^2+2(2)x+c            { c=2^2  }

⇒  x^2+2(2)x+2^2

⇒  (x+2)^2

Therefore , the value of c that makes the expression a perfect square binomial is c=4 .

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3 years ago
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Masteriza [31]

Answer:

<h3>Lucas and Zoe will both next visit their mother on 7th of May at the same time.</h3>

Step-by-step explanation:

<em>First</em><em>,</em><em> </em><em>find </em><em>the </em><em>first</em><em> </em><em>multiple </em><em>that </em><em>can </em><em>be </em><em>divisible </em><em>by </em><em>5</em><em> </em><em>and </em><em>8</em><em>:</em>

<em>4</em><em>0</em><em> </em><em>is </em><em>divisible </em><em>by </em><em>5 </em><em>and </em><em>8</em><em>.</em>

<em>Second</em><em>,</em><em> </em><em>find </em><em>the </em><em>days </em><em>in </em><em>April </em><em>and </em><em>subtract </em><em>7</em><em> </em><em>from </em><em>the </em><em>days:</em>

<em>April</em><em>=</em><em>3</em><em>0</em><em> </em><em>days</em><em>.</em>

<em>3</em><em>0</em><em>-</em><em>7</em><em>=</em><em> </em><em>2</em><em>3</em><em> </em><em>days</em><em>.</em>

<em>Finally,</em><em> </em><em>subtract </em><em>2</em><em>3</em><em> </em><em>from </em><em>4</em><em>0</em><em> </em><em>and </em><em>find </em><em>it </em><em>in </em><em>the </em><em>next </em><em>month</em><em>,</em><em> </em><em>which </em><em>is </em><em>May</em><em>:</em>

<em>4</em><em>0</em><em>-</em><em>2</em><em>3</em><em>=</em><em>1</em><em>7</em>

<em>Therefore</em><em>,</em><em> </em><em>they </em><em>will </em><em>both </em><em>next </em><em>visit </em><em>their </em><em>mother </em><em>on </em><em>the </em><em>7</em><em>t</em><em>h</em><em> </em><em>of </em><em>May.</em>

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