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4 years ago

Debby needed one-third of a cup of water for 1 flower . If she had nine flowers how many cups would she need ?

Mathematics

2 answers:

Ilya [14]4 years ago

5 0

If one flower needs 1/3 of a cup, nine times that amount would need 9/3. Simplified, this would be 3 cups.

Artemon [7]4 years ago

4 0

1/3 per flower so 3 would equal 3/3 or 1 and she was 9 flowers so it would be 3 cups

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Answer:

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Which expression is equivalent to 56? A) 5 × 5 × 5 × 5 × 5 × 5 B) 6 × 6 × 6 × 6 × 6 × 6 × 6 C) 6 × 5 D) 5 6

Sunny_sXe [5.5K]

$\large\text{Hey there!}$

$\large\text{Which expression is equivalent to 56?}$

$\mathsf{5\times5\times5\times5\times5\times5}\\\mathsf{5\times5=25}\\\mathsf{25\times5\times5\times5\times5}\\\mathsf{25\times5=125}\\\mathsf{125\times5\times5\times5}\\\mathsf{125\times5=625}\\\mathsf{625\times5\times5}\\\mathsf{625\times5=3,125}\\\mathsf{3,125\times5=\bf{15,625}}\\\\\\\large\text{Option A. IS NOT your answer since it's too big!}$

$\mathsf{ 6\times6\times6\times6\times6\times6\times6}\\\mathsf{6\times6=36}\\\mathsf{36\times6\times6\times6\times6\times6}\\\mathsf{36\times6=216}\\\mathsf{216\times6\times6\times6\times6}\\\mathsf{216\times6=1,296}\\\mathsf{1,296\times6\times6\times6}\\\mathsf{1,296\times6=7,776}\\\mathsf{7,776\times6\times6}\\\mathsf{7,776\times6= 46,656}\\\mathsf{46,656\times6=\bf{279,936}}\\\\\\\large\text{Option B. IS NOT our answer because its too big!}$

$\mathsf{6\times5=30}\\\\\large\text{Option C. IS NOT your answer because it is a little bit too small!}$

$\mathsf{56=56}\\\\\\\large\text{Option D IS most likely your answer!}$

$\boxed{\boxed{\bf{Thus\ your\ answer\ is: \boxed{\large\text{D) 56}}}}}\checkmark$

$\large\text{Good luck on your assignment and enjoy your day!}\\\\\\\dag\dfrac{\frak{LoveYourselfFirst}}{:)}$

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4 years ago

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iren2701 [21]

Answer:

6 : 8

Step-by-step explanation:

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4 years ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

mixer [17]

Answer:

Step-by-step explanation:

First I want to set up some variables.

S = sum

n = final term

ax = xth term, so a1 is the first term and an is the last one

d = common difference.

There are two formulas to find the sum, if you don't get how they were gotten I'd be happy to explain.

S = (n/2)(a1+an) = (n/2)(2a1+(n-1)d)

So, to find n we use the first one.

S = (n/2)(a1+an)

123 = (n/2)(8+33)

123 = (n/2)41

3 = n/2

6 = n

Now we can find d with the other one

S = (n/2)(2a1+(n-1)d)

123 = (6/2)(2*8+(6-1)d)

123 = 3(16+5d)

41 = 16+5d

25 = 5d

5 = d

so there are six terms and the common difference is 5.

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3 years ago

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3 years ago