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bearhunter [10]
3 years ago
9

1 (9x+2)=4+ 2 (6x+4) 3 3

Mathematics
1 answer:
mote1985 [20]3 years ago
6 0

Answer:i think

Step-by-step explanation:

1(9x+2)=4+2(6x+4)33

=9 x (396 x + 268) + 2 (396 x + 268)

=9 * 396 x x+2 * 396 x+9 * 268 x+2 * 268

=3564 x^2 + 3204 x + 536

(x^2+2)(x+8) (or) (x^3+5)(x+6) (or) (x+2)(x+8)(x+6)

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The ratio of quarters to dimes in a coin collection is 5:3. You add the same number of new quarters as dimes
Anon25 [30]

The ratio of quarters to dimes is not still 5 : 3

<u>Solution:</u>

Given that ratio of quarters to dimes in a coin collection is 5:3 .

You add same number of new quarters as dimes to the collection .  

Need to check if ratio of quarters to dimes is still 5 : 3

As ratio of dimes and quarters is 5 : 3

lets assume initially number of quarters = 5x and number of dimes = 3x.

Now add same number of new quarters as dimes to the collection

Let add "x" number of quarters and "x" number of dimes

So After adding,

Number of quarters = initially number of quarters + added number of quarters = 5x + x = 6x

Number of dimes = initially number of dimes + added number of dimes

= 3x + x = 4x

New ratio of quarters to dimes is 6x : 4x = 3 : 2

So we have seen here ratio get change when same number of new quarters and dimes is added to the collection

Ratio get change from 5 : 3 when same number of new quarters and dimes is added to the collection and new ratio will depend on number of quarters and dimes added to collection.

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4 years ago
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Nat2105 [25]

Answer:

24

Step-by-step explanation:

180-156=24

5 0
3 years ago
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Zielflug [23.3K]
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Solve for k <br> 7k+2m = kr + 4m + 3
professor190 [17]

Answer:

\large\boxed{k=\dfrac{2m+3}{7-r}\ \text{for}\ r\neq7}

Step-by-step explanation:

7k+2m=kr+4m+3\qquad\text{subtract}\ 2m\ \text{from both sides}\\\\7k+2m-2m=kr+4m-2m+3\\\\7k=kr+2m+3\qquad\text{subtract}\ kr\ \text{from both sides}\\\\7k-kr=kr-kr+2m+3\\\\7k-kr=2m+3\qquad\text{distribute}\\\\k(7-r)=2m+3\qquad\text{divide both sides by}\ (7-r)\neq0\\\\k=\dfrac{2m+3}{7-r}

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3 years ago
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