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e-lub [12.9K]
3 years ago
7

Please help! I don't wanna fail!

Mathematics
1 answer:
ahrayia [7]3 years ago
3 0

Answer:

32.55

Step-by-step explanation:

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I ANSWERED A AND B. I ONLY NEED THE ANSWER TO C. PLEASE HELP :(
Andrew [12]

Answer:

1.5

Step-by-step explanation:

Please correct me if I'm wrong. :)

4 0
3 years ago
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The table represents the temperature of a cup of coffee over time.
Zanzabum
<span>Answer: A) exponential, because there is a relatively consistent multiplicative rate of change
</span><span>
temperature thrice. If you compare time=0 until time=30, you will get this table
t(10)- t(0)= 180-200= -20
t(20)-t(10)= 163-180= -17
t(30)-t(20)= 146-163= -17
</span>t(40)-t(30)= 131- 146= -15
t(50)-t(40)= 118-131= -13
t(50)-t(40)= 107-118= -11

There is something weird about the 2nd and 3rd value because it was same -17., but there seems to be a +2 change to the difference. So, the graph would be exponential with a <span>relatively consistent multiplicative rate of change</span>
9 0
3 years ago
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1. determine if the number is written in scientific notation. if not, explain 32x10^4
lozanna [386]
1) 32*10^4 is not written in scientific notation because the first factor is not a number betweem 1 and 10.

3.4 * 10^5 is the same number written in scientific notation.

2) 2.01x10^-5 is written in scientific notation.
5 0
4 years ago
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if you buy 8 chocolate bars at the sale price of .61 cents how many chocolate bars could you have purchased at the regular price
Dahasolnce [82]

If you can buy [email protected] you must have at least 8(61)=488 cents

We get 488/75 = 6.50666...  So at 75 cents a pop, you can buy 6 and have some change left over.

Answer: 6


4 0
4 years ago
Consider the infinite geometric series ∑∞ n=1 -4(1/3)^n-1
erma4kov [3.2K]

a. The series is

\displaystyle\sum_{n=1}^\infty-4\left(\frac13\right)^{n-1}=-4-\frac43-\frac4{3^2}-\frac4{3^3}-\cdots

(first four terms are listed)

b. The series converges because this is a geometric series with r=\dfrac13.

c. Let S_N be the N-th partial sum of the series:

S_N=\displaystyle\sum_{n=1}^N-4\left(\frac13\right)^{n-1}

S_N=-4-\dfrac43-\dfrac4{3^2}-\cdots-\dfrac4{3^{N-1}}

Multiplying both sides by \dfrac13 gives

\dfrac13S_N=-\dfrac43-\dfrac4{3^2}-\dfrac4{3^3}-\cdots-\dfrac4{3^N}

Subtracting this from S_N gives

S_N-\dfrac13S_N=\dfrac23S_N=-4+\dfrac4{3^N}

\implies S_N=-6+\dfrac6{3^N}

As N gets larger and larger (N\to\infty) the rational term converges to 0 and we're left with

\displaystyle\lim_{N\to\infty}S_N=\sum_{n=1}^\infty-4\left(\frac13\right)^{n-1}=-6

8 0
3 years ago
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