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3 years ago

Hailey paid $13 for 1 3/7 kg of sliced salami What was the cost per kilogram of salami

Mathematics

2 answers:

3241004551 [841]3 years ago

6 0

It cost hailey $9.10 for 1 kg of salami

JulijaS [17]3 years ago

4 0

Translate words into numbers.

13/(1 3/7)

First convert to improper fraction.

13/(10/7)

Then, note that division is the same as multiplying by the divisor's reciprocal.

13x7/10=9.1

answer: $9.10 per kilogram

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Answer:

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Step-by-step explanation:

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3 years ago

Bags of sugar come in 3 sizes: Small bag: A 250 g bag costs 95p. Medium bag: A 500 g bag costs £1.80. Large bag: A 3 kg bag cost

azamat

Answer:

see below

Step-by-step explanation:

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3 years ago

What is the sum of 1/6 , 2/3 , and 1/4

babymother [125]

Answer:

13/12

Step-by-step explanation:

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8 0

3 years ago

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Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

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Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago

Please help will mark brainliest

atroni [7]

1,4,5,8

Hope this helps! <3

(Im so sorry if its wrong i havent done these in a while but in 99% sure)

7 0

4 years ago

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