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Tomtit [17]
3 years ago
9

What times what equals 19

Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
5 0
19 is prime meaning that only 1 and itself can multiply to get that number (excluding decimals and fractions)
so the answer would be 1 times 19

or if you want to multiply by a fraction to divide you could do
38 times 1/2=19
Vlada [557]3 years ago
3 0
No whole numbers with evenly equal 19. So the closest would be 3.8 x 5 or 4 x 4.75.
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How to solve number 44
solniwko [45]
OK. 

Let's say a member and a non member each visit the garden ' V ' times.

The non-member's cost for each visit is  $6 .
The non member's cost for ' V ' visits is  6 V .
His total cost for the year is  6 V .

The member's cost for each visit is $3.
The member's cost for ' V ' visits is  3V .
His total cost for the year is  3V  +  the  $24 to join.

We want to know what ' V ' is  (how many times each one can visit)
if their total costs are the same.

So let's just write an equation that SAYS their costs are the same,
and see what ' V ' turns out to be.

             Non-member's cost for the year =  Member's cost for the year

                                                            6 V  =  3 V + 24

Subtract  3V  from each side:          3 V  =            24

Divide each side by  3 :                       V  =              8  .

-- If they both visit the garden 1, 2, 3, 4, 5, 6, or 7 times in the year,
the member will spend MORE than the non member.

-- If they both visit the garden 8 times in the year,
they'll both spend the same amount.  ($48)

-- If they both visit the garden MORE than 8 times in the year,
the member will spend LESS than the non-member.
______________________________________________ 

That was the algebra way to do it.

Now here is the cheap, sleazy, logical, easy way to do it:

The non-member spends (6 - 3) = $3 MORE than the member for each visit ?

After how many visits does the $3 more each time add up to the $24 that
it cost the member to join for the year ?

           $24 / $3  =  8 visits  .
5 0
4 years ago
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