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matrenka [14]
3 years ago
13

Select all the expressions that represent the large rectangle's total area can't find the answer please help

Mathematics
1 answer:
koban [17]3 years ago
8 0
Where are the expressions. you didn’t attach anything
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you are playing the integer game with your friends. you have drawn a -2 and -3 from the deck. when you add the numbers on these
ahrayia [7]

Answer:

Bb

Step-by-step explanation:

Njkkmknb

4 0
3 years ago
Which is a possible way to rewrite the equation y=3x+3b to solve for b
alex41 [277]
3b=y+3x 


To solve for B you need to put B before the equals sign 
6 0
3 years ago
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Question 3<br> And question 4 pleaseeee help!!
Verizon [17]

Answer:

see pdf

Step-by-step explanation:

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4 0
3 years ago
If r and s are positive integers, is \small \frac{r}{s} an integer? (1) Every factor of s is also a factor of r. (2) Every prime
Yuri [45]

Answer:

<em>If statement(1) holds true, it is correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em>If statement(2) holds true, it is not necessarily correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em></em>

Step-by-step explanation:

Given two positive integers r and s.

To check whether \small \frac{r}{s} is an integer:

Condition (1):

Every factor of s is also a factor of r.

r \geq s

Let us consider an example:

s = 5^2 \cdot 2\\r = 5^3 \cdot 2^2

\dfrac{r}{s} = \dfrac{5^3\cdot2^2}{5^2\cdot2} = 10

which is an integer.

Actually, in this situation s is a factor of r.

Condition 2:

Every prime factor of <em>s</em> is also a prime factor of <em>r</em>.

(But the powers of prime factors need not be equal as we are not given the conditions related to powers of prime factors.)

Let

r = 2^2\cdot 5\\s =2^4\cdot 5

\dfrac{r}{s} = \dfrac{2^3\cdot5}{2^4\cdot5} = \dfrac{1}{2}

which is not an integer.

So, the answer is:

<em>If statement(1) holds true, it is correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em>If statement(2) holds true, it is not necessarily correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em></em>

8 0
4 years ago
Find three consecutive even integers such that 3 times the first is 26 less than twice the sum of the last two
Dimas [21]
Let's say our first integer is "a".

how to get the next consecutive EVEN integer?  well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".

and the next after that, will then be (a + 2) + 2, or "a + 4".

so those are are 3 integers, a           a + 2           a+4

notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.

2 + 2 is 4, or 8 + 2 is 10  some even ones

3 + 2 is 5, or 13 + 2 is 15, some odd ones

\bf \stackrel{\textit{3 times the first}}{3a}~~=~~\stackrel{\textit{26 less than twice the sum of the others}}{2[~(a+2)+(a+4)~]~~~-26}&#10;\\\\\\&#10;3a=2[~2a+6~]-26\implies 3a=4a+12-26\implies 3a=4a-14&#10;\\\\\\&#10;0=a-14\implies 14=a

what are the other two consecutive integers?  well, a + 2 and a + 4.
4 0
3 years ago
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