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2 years ago

What is the answer to 7y + 5 + (3 - y)?

Mathematics

2 answers:

Pachacha [2.7K]2 years ago

6 0

Answer:

6y +8

Step-by-step explanation:

in order to solve this, take note of the bracket and combine the like terms.

solution

7y + 5 + (3 - y)

7y + 5 +3 -y

7y-y + 5+3

6y +8

Nataliya [291]2 years ago

4 0

Answer with factoring: 2(3y+4)

Answer with simplifying: 6y+8

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There are $1.95 worth of nickels and quarters in a piggy bank there are three more nickels than quarters how many of each are in

Gennadij [26K]

Answer:

6 quarters and 9 nickels

Step-by-step explanation:

6 quarters is equal to $1.50. 6 nickels is equal to $0.30. Those together equal $1.80, then add the three nickels bringing your total to $1.95.

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2 years ago

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Sidana [21]

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2 years ago

[20 pts] evaluate the logarithm log(6)1/36, show work pls!

Delvig [45]

Answer:

$\large\boxed{\log_6\dfrac{1}{36}=-2}$

Step-by-step explanation:

$\text{We know:}\\\\\log_ab=c\iff a^c=b\\\\\log_6\dfrac{1}{36}\qquad\text{use}\ a^{-1}=\dfrac{1}{a}\\\\=\log_636^{-1}=\log_6(6^2)^{-1}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\log_66^{(2)(-1)}=\log_66^{-2}\qquad\text{use}\log_ab^n=n\log_ab\\\\=-2\log_66\qquad\text{use}\ \log_aa=1\\\\=-2(1)=-2$

$\log_6\dfrac{1}{36}=-2\ \text{because}\ 6^{-2}=\dfrac{1}{6^2}=\dfrac{1}{36}$

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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.

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

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