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2 years ago

NEED HELP WITH THIS

Mathematics

1 answer:

Kaylis [27]2 years ago

5 0

Answer:

Step-by-step explanation:

think of this as a system of equations where the colors just represent a variable like x or y

I'll just arbitrarily set colors to some variable letter so we have

purple = x

green =y

blue =z

red = a

yellow = b

so we can now rewrite the 4 known equations of the system :)

Eq. 1) x+y = 13

Eq. 2) 2y+2z=18

Eq. 3) 2x+3y=31

Eq. 4) z + 4a = 12

Eq. 5) a * y = b

so solve the 1st eq for y => y =13-x then plug that into the 3rd eq and slove for x => 2x + 3(13-x) =31

2x + 39 -3x = 31

2x-3x = 31 - 39

-x = -8

x=8

plug 8 into Eq. 1 for x and solve for y

8+y = 13

y=5

plug y into Eq 1 and solve for z

2(5) +2z = 18

10 +2z = 18

2z = 8

z = 4

plug z into E1 4 and solve for a

4 + 4a = 12

4a = 8

a = 2

plug a and y into Eq 5 and solve b

2 * 5 = b

10=b

go to the drawing and plug in each of the color values that we put as variable letters above and get an A+ on your homework.. :)

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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.

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

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