*The sum of two consecutive numbers is n+1. What are the numbers?

Mathematics

2 answers:

Verizon [17]3 years ago

6 0

The answer will be 1 and n which make the sum equal n+1. Hope it help!

Nostrana [21]3 years ago

3 0

Answer:

the numbers are n and 1

Step-by-step explanation:

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You need 150g of flour for 12 cakes how much flour do you need

Vaselesa [24]

Answer:

12.5g for 1 cake, 150g for 12 cakes

Step-by-step explanation:

Because you need 150g for 12 cakes so 1 cake will take:

150 / 12 = 12.5g

So, You need 12.5g for 1 cake

Hope this help :3

8 0

3 years ago

Read 2 more answers

1. The current ratio (assets/liabilities) of company X is 3.1. Given that the current assets are $186000, find the current liabi

katrin [286]

<span>1. The current ratio (assets/liabilities) of company X is 3.1. Given that the current assets are $186000, find the current liabilities:

Assets = A
Liabilities = L

A / L = 3.1 => L = A / 3.1

A = 186000 => L = 18600 / 3.1 = 6000

Answer: $ 6000

2. The board of directors determines that the current ratio must never be below 2.6. What is the maximum amount that the company can borrow?

A/L ≥ 2.6

=> L ≤ A / 2.6

=> L ≤ 186000 / 2.6

L ≤ 71538.46

Answer: $71538.46
</span>

6 0

3 years ago

Find the sum of the geometric series 512+256+ . . .+4

mario62 [17]

$\bf 512~~,~~\stackrel{512\cdot \frac{1}{2}}{256}~~,~~...4$

so, as you can see above, the common ratio r = 1/2, now, what term is +4 anyway?

$\bf n^{th}\textit{ term of a geometric sequence}\\\\a_n=a_1\cdot r^{n-1}\qquad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\a_n=+4\end{cases}$

$\bf 4=512\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{4}{512}=\left( \cfrac{1}{2} \right)^{n-1}\\\\\\\cfrac{1}{128}=\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{1}{2^7}=\left( \cfrac{1}{2} \right)^{n-1}\implies 2^{-7}=\left( 2^{-1}\right)^{n-1}\\\\\\(2^{-1})^7=(2^{-1})^{n-1}\implies 7=n-1\implies \boxed{8=n}$

so is the 8th term, then, let's find the Sum of the first 8 terms.

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\n=8\end{cases}$

$\bf S_8=512\left[ \cfrac{1-\left( \frac{1}{2} \right)^8}{1-\frac{1}{2}} \right]\implies S_8=512\left(\cfrac{1-\frac{1}{256}}{\frac{1}{2}} \right)\implies S_8=512\left(\cfrac{\frac{255}{256}}{\frac{1}{2}} \right)\\\\\\S_8=512\cdot \cfrac{255}{128}\implies S_8=1020$