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DaniilM [7]
3 years ago
13

Write the sum using summation notation, assuming the suggested pattern continues. -9 - 3 + 3 + 9 + ... + 81

Mathematics
1 answer:
noname [10]3 years ago
8 0

Answer:

Option A:        $ \sum_{n = 1}^{15} {\textbf{- 9 + 6 n}} $

Step-by-step explanation:

We are given with the series - 9  - 3  + 3   +  9 +  . . . .  +  81.

Note that the second term is obtained by adding 6 to the first term.

Each consecutive term is obtained by adding 6 to its previous term.

Therefore, we should be adding six two times to get the third term from the first term.

Putting it Mathematically, we get: - 9 + 6n

This gives all the terms of the sequence. Since, we have to add all the terms we take the summation.

Also, note that 81 is the 15 th term.

Therefore, - 9  - 3  + 3   +  9 +  . . . .  +  81 =    $ \sum_{n = 1}^{15} {- 9 + 6 n} $

Hence, the answer.

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Factor using a GCF:<br> ​ 16xy2+28xy+8y
Lorico [155]

Answer:

4y( 4xy + 7x +2)

Step-by-step explanation:

16xy^2+28xy+8y

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The common factors are 2*2*y = 4y

2*2*2*2*x*y*y + 2*2*7 *x*y+2*2*2*y

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3 years ago
Find m1 and m2 <br><br>m2 (×10+10)° <br>m1 (3×+4)°​
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Step-by-step explanation:

5 0
2 years ago
Help!! Answer !!! about to run out of time in test!!
Gemiola [76]

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the solution ~

Let's find the measure of hypotenuse first, by using Pythagoras theorem ;

\qquad \sf  \dashrightarrow \: h {}^{2}  =  {8}^{2}  +  {6}^{2}

\qquad \sf  \dashrightarrow \: h {}^{2}  =  {36}^{}  +  {64}^{}

\qquad \sf  \dashrightarrow \: h {}^{2}  = 100

\qquad \sf  \dashrightarrow \: h {}^{}  =  \sqrt{100}

\qquad \sf  \dashrightarrow \: h {}^{}  =  {10}

Now, let's find the asked values ~

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{opposite \: side}{hypotenuse}

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{6}{10}

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{3}{5}   \: or \: 0.6 \: units

For Cos y :

\qquad \sf  \dashrightarrow \:  \cos(y) =  \dfrac{adjcant \: side}{hypotenuse}

\qquad \sf  \dashrightarrow \:  \cos(y)   = \dfrac{6}{10}

\qquad \sf  \dashrightarrow \:  \cos(y)   = \dfrac{3}{5}  \: or \: 0.6 \: units

As we can see that both sin x and Cos y have equal values, therefore The required relationships is equality.

I.e Sin x = Cos y

Hope it helps ~

3 0
2 years ago
Is this correct or no?
Tanya [424]
Yes.??? dont really Know sorry
4 0
3 years ago
Read 2 more answers
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