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jekas [21]
2 years ago
11

Which expressions are equivalent to

Mathematics
2 answers:
klio [65]2 years ago
7 0

Answer:

Tired

Step-by-step explanation:

riadik2000 [5.3K]2 years ago
4 0
<h3>Given ↓</h3>
  • The expression 6+12x

<h3>To Find ↓</h3>
  • Which expressions are equivalent to 6+12x?

<h3>Calculations ↓</h3>

  Let's see the other options :

7(1+2x)

Use the distributive property :

7+14x

Is 7+14x equivalent to 6+12x? No.

The expression equivalent to 6+12x is :

6(1+2x) ← divide 6+12x by 6 and put it outside the parentheses

hope helpful ~

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Emma is planning a surprise party. she is considering.......
Ann [662]

Answer:


Step-by-step explanation:

what is she considering? I would need more information to complete the probolem

5 0
3 years ago
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What is the additive inverse of 7/2 ?<br> A) −2/7<br> B) −7/2<br> C) 2/7<br> D) 7/2
snow_lady [41]
The additive inverse of 7/2 would be -7/2 (option B) Since its. opposite !

A)is the additive inverse of C) ; C) is the additive inverse of A) (not your answer)

B. is the additive inverse for this problem (7/2) so thats how we found that this is your answer (answer)

D is the same thing for the problem (7/2) (not your answer)

7/2 isnt the additive inverse for 7/2 because they are the SAME not the opposite.

+ = - (Positive = negative)

- = + (negative = positive)
6 0
3 years ago
Find the Absolute Value of: |-16+9|
musickatia [10]

Answer:

7

Step-by-step explanation:

|-16+9|

Calculate the number inside the absolute value

|-7|

Now take the absolute value

7

5 0
3 years ago
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Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.
Lilit [14]

For starters,

\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k

Consider the nth partial sum, denoted by S_n:

S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n

Multiply both sides by \frac34:

\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}

Subtract S_n from this:

\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)

Solve for S_n:

-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)

S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)

Now as n\to\infty, the exponential term will converge to 0, since r^n\to0 if 0. This leaves us with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}

8 0
3 years ago
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What's the anewer to this problem..?
valentina_108 [34]
21x(9)=9x21
21y= 9x21
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Y=9x1
Y=9
4 0
3 years ago
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