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Sergeu [11.5K]
4 years ago
11

How many three-digit numbers are there such that the first two digits are even and the last digit is odd? Answer is a whole numb

er.
Mathematics
1 answer:
Genrish500 [490]4 years ago
5 0
221
223
225
227
229
241
243
245
247
249
261
263
265
267
269
281
283
285
287
289
421
423
425
427
429
441
443
445
447
449
461
463
465
467
469
481
483
485
487
489
621
623
625
627
629
641
643
645
647
649
661
663
665
667
669
681
683
685
687
689
821
823
825
827
829
841
843
845
847
849
861
863
865
867
869
881
883
885
887
889
80 three-digit numbers 
(just to make sure, check)

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koban [17]

Answer:

The simplified form of -6.3x+14 and 1.5x-6 is -4.8x+8

Step-by-step explanation:

We have to simplify the following

-6.3x+14 and 1.5x-6

it can be written as:

=(-6.3x+14) + (1.5x-6)

Adding the like terms

=(-6.3x+1.5x)+(14-6)

= (-4.8x)+(8)

= -4.8x+8

So, the simplified form of -6.3x+14 and 1.5x-6 is -4.8x+8

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4 years ago
(12a^(2) - 3)/(2)*(2a + 1)^(-2)*((6)/((2a + 1)))^(-1)
Elena L [17]

Answer:

\frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\

Step-by-step explanation:

\frac{(12a^2-3)}{2*(2a+1)^{-2}}*\frac{6}{(2a+1)^{-1} }\\ = \frac{(12a^2-3)}{2 * \frac{1}{(2a+1)^{2} }} * \frac{6}{\frac{1}{2a+1}  }\\ =\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}\\=\frac{(12a^2-3)(2a+1)^{2}}{2} * \frac{6}{2a+1}

= \frac{(12a^2-3)(4a^{2} + 1 + 2 (2a)(1))}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{2} * \frac{6}{2a+1}\\= \frac{(12a^2-3)(4a^{2} + 1 + 4a)}{1} * \frac{3}{2a+1}\\= \frac{3(12a^2-3)(4a^{2} + 1 + 4a)} {2a+1}\\\\= \frac{3(12a^2(4a^{2} + 1 + 4a) - 3 (4a^{2}+1+4a)} {2a+1}\\= \frac{3(48a^{4}+12a^{2}+48a^{3}  - 12a^{2} -12 - 4a)} {2a+1}\\= \frac{144a^{4}+36a^{2}+144a^{3}  - 36a^{2} - 36 - 12a)} {2a+1}

= \frac{144a^{4}+36a^{2}+144a^{3}  - 36a^{2} - 36 - 12a} {2a+1}\\= \frac{144a^{4}+144a^{3} - 36 - 12a} {2a+1}\\

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Anyone know the answer too this m-43=14
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Answer:

1:5

Step-by-step explanation:

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