Just multiply the factor (5) by each of the phrases in the parenthesis. So that 5*8d will be added to 5*7e and added to 5*3 to get:
40d + 35e + 15
Good luck.
2007-05-02 16:09:22
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answer #1
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answered by ¼ + ½ = ¾ 3
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The Distributive Property says
a*(b+c)= a*b + a*c
where a, b, and c are anything you could ever want to multiply. In words, you can think
I can either add first and then multiply, or multiply first and then add.
The only caveat is that when you multiply first and then add, you have to make sure you multiply EVERYTHING by the a, and then add all the resulting products.
As an example, consider
2*(3+4)
You can add first, getting 3+4=7
Then multiply, 2*7=14.
Or, you can multiply first, 2*3=6 and 2*4=8
Then add, 6+8=14
Both ways, same thing. The second way seems stupid, right? You have to do three things instead of the two things in the first way.
But then you have your problem. There's no way to add 8d and 7e and 3! So your only resort is to do the second way.
I'll do some of it, but let you fill in some of it too...
1) 5*8d = ___
2) 5*7e = ___
3) 5*3 = ___
So that's multiply first. Now we add the results of 1, 2, and 3 together, combining like terms if possible.
And if anyone's ever talked to you about "factoring" they're really just using the distributive property in reverse...
Hope that helps, and best of luck!
2007-05-02 16:16:29
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answer #2
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answered by ihoj5 1
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just multiply each term by 5
5 x 8d = 40d
5 x 7e = 35e
5 x 3 = 15
40d + 35e+ 15
just "distribute" the 5 into the terms of the equation.
2007-05-02 16:09:45
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answer #3
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answered by smeagol_jr 4
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5 distributes to every term
5(8d) + 5(7e) + 5(3)
40d + 35e + 15
:)
2007-05-02 16:08:50
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answer #4
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answered by Anonymous
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If you can't do this by yourself, get a job as a camal driver. PS. dont forget the 5 multiplies ALL coefficients in all term in the ( )
2007-05-02 16:09:51
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answer #5
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answered by cattbarf 7
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40d + 35e + 15
2007-05-02 16:08:50
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answer #6
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answered by hookemhornsfan1991 2
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just multiply everything by 5
40d + 35e + 15
thats as far as you can go because e and d are different variables
2007-05-02 16:09:54
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answer #7
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answered by Anonymous
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considering that ten tenths of something make up an finished (10 dimes make up a greenback, etc.), there might want to be TEN MILLION tenths in a million. someone suggested to divide a million through 10, yet this merely tells you what one-10th of one million might want to be, not what number 10th instruments are in a unit of one million.
2016-12-05 06:40:52
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answer #8
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answered by ? 4
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