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maw [93]
3 years ago
12

Question 5

Mathematics
1 answer:
yarga [219]3 years ago
5 0

The answer choice 12,357 is correct. Since the numbers are ascending from lowest to greatest, the number couldn’t possibly be any smaller then this certain answer choice. So the correct answer to this question is 12,357.

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When asked to write the domain for the equation y= |x - 6| + 3, Mary wrote the Domain is y < 3. What mistake(s) did Mary make
castortr0y [4]

Answer:

Ok, the domain is the set of values that we can input in a function.

In this case, we have:

y = Ix - 6I + 3.

Notice that there is no restriction here, x can actually take any value, then the domain will be the set of all real numbers.

The correct domain is x, x ∈ R

Now, if we had (for example) something like:

y = Ix - 6I < 3

Now we have a restriction in the domain because we can not have y equal or larger than 3.

To find the domain, we can break the absolute value:

Ix - 6I < 3

is equivalent to:

-3 < x - 6 < 3

now let's add 6 in each side.

-3 + 6 < x - 6 + 6 < 3 + 6

3 < x < 9

That will be the domain in that case.

6 0
3 years ago
Fiona wrote the linear equation y = x – 5. when henry wrote his equation, they discovered that his equation had all the same sol
djverab [1.8K]
I would go with A but I am not 100% sure so you may want to check however you can.
6 0
3 years ago
Read 2 more answers
If five times four is thirty-three, what will the fourth of twenty be?
Zarrin [17]

Answer:

The fourth of twenty is 5

Step-by-step explanation:

The 4th of twenty implies that we divide the number twenty into 4 place, let's us divide it 4

20/4= 5

This problem bothers on on division of numbers and it requires proper understanding of words/terms associated with it.

5 0
3 years ago
4y − 3x − 2y 11 x − 2 x
Inessa05 [86]
By rearranging it:
-3x-2x-22xy+4y
=-5x-22xy+4y answer
4 0
3 years ago
If a positive integer is equal to the following product: 25b3c425b3c4, where b and c are distinct prime numbers greater than 2,
Vlad [161]

Answer: 64 distinct even factors

Step-by-step explanation:

let the 2 distinct numbers be b and c and the integer is expected to be 25b3c425b3c4

since b, c > 2 and prime numbers,

potential options include 3, 5, and 7

hence the likelihoods are (b = 3, c = 5), (b = 5, c = 3), (b = 3, c = 7), (b = 7, c = 3), (b = 5, c = 7), (b = 7, c = 5)

Possibility 1 (b = 3, c = 5)

integer is now 253354253354

the distinct even factors = 2, 202, 262, 1934, 19802, 26462, 195334, 253354, 2000002, 2594062, 19148534, 25588754, 262000262, 1934001934, 2508457954, 253354253354

number of distinct even factors = 16

Possibility 2 (b = 5, c = 3)

integer is now 255334255334

the distinct even factors = 2, 86, 202, 5938, 8686, 19802, 253354, 599738, 851486, 2000002, 25788734, 58792138, 25588754, 86000086, 2528061934, 1934001934, 5938005938, 253354253354

number of distinct even factors = 18

Possibility 3 (b = 3, c = 7)

integer is now 253374253374

the distinct even factors = 2, 6, 22, 66, 202, 242, 606, 698, 726, 2094, 2222, 6666, 7678, 19802, 23034, 24442, 59406, 70498, 73326, 84458, 211494, 217822, 253374, 653466, 775478, 2000002, 2326434, 2396042, 6000006, 6910898, 7188126, 8530258, 20732694, 22000022, 25590774, 66000066, 76019878, 228059634, 242000242, 698000698, 726000726, 836218658, 2094002094, 2508655974, 7678007678, 23034023034, 84458084458,253354253354

number of distinct even factors = 48

Possibility 4 (b = 7, c = 3)

integer is now 257334257334

the distinct even factors = 2, 6, 14, 22, 42, 66, 154, 202, 462, 606, 1114, 1414, 2222, 3342, 4242, 6666, 7798, 12254, 15554, 19802, 23394, 36762, 46662, 59406, 85778, 112514, 138614, 217822, 257334, 337542, 415842, 653466, 787598, 1237654, 1524754, 2000002, 2362794, 3712962, 4574262, 6000006, 8663578, 11029714, 14000014, 22000022, 25990734, 33089142, 42000042, 66000066, 77207998, 121326854, 154000154, 231623994, 363980562, 462000462, 849287978, 1114001114, 2547863934, 3342003342, 7798007798, 12254012254, 23394023394, 36762036762, 85778085778, 257334257334

number of distinct even factors = 64

Possibility 5 (b = 5, c = 7)

integer is now 255374255374

the distinct even factors = 2, 14, 34, 58, 74, 202, 238, 406, 518, 986, 1258, 1414, 2146, 3434, 5858, 6902, 7474, 8806, 15022, 19802, 24038, 36482, 41006, 52318, 99586, 127058, 138614, 216746, 255374, 336634, 574258, 697102, 732674, 889406, 1517222, 2000002, 2356438, 3684682, 4019806, 5128718, 9762386, 12455458, 14000014, 21247546, 25792774, 34000034, 58000058, 68336702, 74000074, 87188206, 148732822, 238000238, 361208282, 406000406, 518000518, 986000986, 1258001258, 2146002146, 2528457974, 6902006902, 8806008806, 15022015022, 36482036482, 255374255374

number of distinct even factors = 64

Possibility 6 (b = 7, c = 5)

integer is now 257354257354

the distinct even factors = 2, 202, 19802, 257354, 2000002, 25992754, 2548061954, 257354257354

number of distinct even factors = 8

From the six possibilities, the highest number of likely distinct even factors is 64

7 0
3 years ago
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