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harkovskaia [24]
3 years ago
14

Can someone please help me fill this out I generally need help

Mathematics
1 answer:
Lapatulllka [165]3 years ago
3 0

Answer:

12

Step-by-step explanation:

ill take 12 dollars

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-x - 4y = 16 find value of y when x equals -8
pashok25 [27]
The answer is -2......
8 0
3 years ago
Read 2 more answers
I need help answering all of these. My teacher thinks that we should know this even though he hasn´t taught us any of this. plea
bearhunter [10]

Answer:

Step-by-step explanation:

x/2=3.5

x=3.5*2

x=7

3 0
3 years ago
HELP!!<br> This is for my geometry exam and i don’t understand <br> please help
rusak2 [61]

Answer:

Step-by-step explanation:

Sin 58° = 0.848 = BC/13

So

BC = 0.848*13 = 11.024

4 0
3 years ago
brittany wants to make cookies at her school one batch of dough makes 2 dozen cookies she has 68 students at her school is it re
Anna [14]
Yes, it is reasonable for Brittany to think that she will need more than 2 batches of cookies to feed all of the people at her school.


One batch of dough makes 2 dozen cookies, or 24 cookies.
This means that two batches of dough makes 4 dozen cookies, or 48 cookies.

If Brittany only brought 48 cookies to school, some kids wouldn't get cookies.  So, we know that Brittany should probably bake more than 2 batches of cookies.
3 0
4 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

7 0
4 years ago
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