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3 years ago

one liter of strawberries weight about 4 kilograms about hkw many liters of strawberries would weight 400 grams

1 answer:

4 0

It says the in every 1 liter of strawberries weighs 4 kilograms.
Find how many liters of strawberries are there in 400 kilograms.
=> to get the answer, divide the 400 by 4
=> 400 / 4
=> 100 liters of strawberries
=> to check if our answer is correct.
To this formula
=> answer x divisor
=> 100 x 4 = 400
Thus, there are 100 liters of strawberries and weighs for about 400 kilograms all in all.

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Two pairs of statements about Jeff's housing expenditures are given below. How can both pairs of statements be true?

Irina18 [472]

Explanation:

Pair 1 is true if Jeff's monthly income is $600/20% = $3,000.

Pair 2 is true if Jeff's monthly income is $1200/10% = $12,000.

Both pairs can be true if Jeff's monthly income increased by a factor of 4 in the 20 years from 1990 to 2010.

Obviously, Jeff spent more on housing in 2010. (Fortunately for Jeff, that larger expenditure was a smaller fraction of his income.)

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2 years ago

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Find the slope of the line that passes through (5,12) and (9,7)

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1 year ago

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.

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3 years ago