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zepelin [54]
2 years ago
9

CAN SOMEBODY PLEASE DO THIS

Mathematics
1 answer:
lesya [120]2 years ago
6 0
A ?
Might wanna double check though
You might be interested in
Find the product. Simplify your answer.<br> -5 • 9t<br> 7s 7
Sauron [17]

Answer:

-13            

 ——— = -0.20635  

 63            

Step-by-step explanation:

Step  1  :

           7

Simplify   —

           9

Equation at the end of step  1  :

 4    7

 — -  —

 7    9

Step  2  :

           4

Simplify   —

           7

Equation at the end of step  2  :

 4    7

 — -  —

 7    9

Step  3  :

Calculating the Least Common Multiple :

3.1    Find the Least Common Multiple

     The left denominator is :       7  

     The right denominator is :       9  

       Number of times each prime factor

       appears in the factorization of:

Prime  

Factor   Left  

Denominator   Right  

Denominator   L.C.M = Max  

{Left,Right}  

7 1 0 1

3 0 2 2

Product of all  

Prime Factors  7 9 63

     Least Common Multiple:

     63  

Calculating Multipliers :

3.2    Calculate multipliers for the two fractions

   Denote the Least Common Multiple by  L.C.M  

   Denote the Left Multiplier by  Left_M  

   Denote the Right Multiplier by  Right_M  

   Denote the Left Deniminator by  L_Deno  

   Denote the Right Multiplier by  R_Deno  

  Left_M = L.C.M / L_Deno = 9

  Right_M = L.C.M / R_Deno = 7

Making Equivalent Fractions :

3.3      Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example :  1/2   and  2/4  are equivalent,  y/(y+1)2   and  (y2+y)/(y+1)3  are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

  L. Mult. • L. Num.      4 • 9

  ——————————————————  =   —————

        L.C.M              63  

  R. Mult. • R. Num.      7 • 7

  ——————————————————  =   —————

        L.C.M              63  

Adding fractions that have a common denominator :

3.4       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

4 • 9 - (7 • 7)     -13

———————————————  =  ———

      63            63  

Final result :

 -13            

 ——— = -0.20635  

 63            

8 0
3 years ago
Read 2 more answers
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
3 years ago
PLEASE ANSWER ALOT OF POINTS
tatyana61 [14]

Answer:

Part A 1 solution

Part B 1,4

5 0
2 years ago
Read 2 more answers
Divide.<br><br><br> Express your answer in lowest terms.<br><br><br> 1/8 Divided by 4 =
Yuliya22 [10]
\sf\dfrac{1}{8}\div4

Find the reciprocal of 4 and multiply.

\sf\dfrac{1}{8}\times\dfrac{1}{4}

Multiply the numerators and denominators together:

\boxed{\sf\dfrac{1}{32}}
5 0
3 years ago
Read 2 more answers
The sum of 6c - 5 and the opposite of 6c<br> (6c - 5) + (-6c)
Julli [10]

Answer:

There are no real solutions for the given question (6c - 5) + (-6c)

Step-by-step explanation:

Solution for (6c-5) + (-6c)

Simplifying ,

(6c + -5) + (-6c) = 0

Reordering the given terms:

(-5 + 6c) + (-6c) = 0

Remove "parenthesis" around (-5 + 6c)

-5 + 6c + (-6c) = 0

Combining like terms it becomes 6c + (-6c) = 0

-5 + 0 = 0

-5 = 0

Solving

-5 = 0

we could not find the solution for the given (6c - 5) + (-6c).

The given equation is not a valid one, the "left hand side" and "right hand side" are unequal, and therefore there is no solution.

5 0
2 years ago
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