5sqrt8 + 2sqrt18 =
5*8^¹/2 + 2*18^¹/2 =
5*[(4*2)^¹/2] + 2*[(9*2)^¹/2] =
5*[(4^¹/2)*(2^¹/2)] + 2*[(9^¹/2)*(2^¹/2)] =
5*[2*(2^¹/2)] + 2*[3*(2^¹/2)] =
5*2*(2^¹/2) + 2*3*(2^¹/2) =
10*(2^¹/2) + 6*(2^¹/2) =
= 16*(2^¹/2)
₢
2006-12-02 05:47:16
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answer #1
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answered by Luiz S 7
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For those with 7 digits, replace it to this: a million. grab a calculator. 2. Key contained in the first 3 digits of your living house telephone volume (no longer the realm code) 3. Multiply with the aid of 80 4. upload a million 5. Multiply with the aid of 250 6. upload the 4 digit # on the proper of your telephone volume 7. upload the 4 digit # on the proper of your telephone volume lower back. 8. Subtract 250 9. Divide volume with the aid of two If it would not supply you the outcomes you want, you may want to nicely be doing something incorrect. be effective to enter = after each step.
2016-11-23 12:47:51
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answer #2
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answered by fette 4
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You have to treat sqrt(2) as a variable.
In the above case, it becomes
10 sqrt(2) + 6 sqrt(2) = 16 sqrt(2)
The same way that
10x + 6x = 16x
Since sqrt(2) is in its simplest form while staying exact, you're not allowed to touch it, but you can combine it as a like term.
2006-12-02 05:40:15
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answer #3
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answered by Puggy 7
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5√(8) + 2√(18)
= 5(√4)(√2) + 2(√9)(√2)
= (5)(2)(√2) + (2)(3)(√2)
= 10√2 + 6√2
If you make √2 into x...
= 10x + 6x --- And you probably know that this...
= 16x
Therefore...
= 10√2 + 6√2
= 16√2
Hope that helped!
2006-12-02 06:04:44
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answer #4
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answered by Anonymous
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5sqrt8 + 2sqrt18
5* sqrt4 *sqrt 2 +2 * sqrt9 *sqrt2
=5*2*sqrt2 +2*3*sqrt2
= 10 sqrt2 +6sqrt2 = 16sqrt2
As shown, you multiply.
2006-12-02 05:45:38
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answer #5
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answered by ironduke8159 7
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5sqrt(8) + 2sqrt(18)
Minute detail follows:
5sqrt(4*2) + 2sqrt(9*2)
5sqrt(4)sqrt(2) + 2sqrt(9)sqrt(2)
5*2sqrt(2) + 2*3sqrt(2)
10sqrt(2) + 6sqrt(2)
16sqrt(2)
2006-12-02 05:42:22
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answer #6
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answered by Anonymous
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5sq.rt8=10sq.rt2
2sq.rt.18=6sq.rt2
so adding
it is 16sq.rt.2
2006-12-02 05:44:09
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answer #7
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answered by raj 7
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multiply i already told you
2006-12-02 05:42:41
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answer #8
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answered by m0rph0s1s 2
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