11. A line with a slope of - 1 passes through points at (2, 3) and (5. y). Find the value of y.

How many different answers can you get by grouping differently? Add parentheses to the expression 7+3•4+2 to create new expressi

DiKsa [7]

Answer: The different possible solutions are 42, 21, 31 and 80.

Explanation:

The given expression is,

$7+3\cdot 4+2$

Case 1.

$(7+3)\cdot 4+2=10\cdot 4+2$

$(7+3)\cdot 4+2=40+2$

$(7+3)\cdot 4+2=42$

Case 2.

$7+(3\cdot 4)+2=7+12+2$

$7+(3\cdot 4)+2=21$

Case 3.

$7+3\cdot (4+2)=7+3\cdot 8$

$7+3\cdot (4+2)=7+24$

$7+3\cdot (4+2)=31$

Case 4.

$(7+3)\cdot (4+2)=10\cdot 8$

$(7+3)\cdot (4+2)=80$

8 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.

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

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