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liberstina [14]
3 years ago
8

Help me someone!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Julli [10]3 years ago
7 0
Substitute answers in, a = 3 and b = 2.

(3^2 - 2^2)/(3+2)
= (9-4)/5
= 5/5
= 1
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Anna71 [15]
The second one the unit rate is 2.05 i think
5 0
3 years ago
If the sin 30° is 1 over 2, then the cos ____° = _____. (4 points) 60° ; 1 over 2 30° ; square root 2 over 2 60° ; square root 3
umka21 [38]

Step-by-step explanation:

\sin 30 \degree =  \cos 60 \degree = \frac{1}{2}  \\

5 0
3 years ago
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For the following telescoping series, find a formula for the nth term of the sequence of partial sums
gtnhenbr [62]

I'm guessing the sum is supposed to be

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}

Split the summand into partial fractions:

\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}

1=a(5k+4)+b(5k-1)

If k=-\frac45, then

1=b(-4-1)\implies b=-\frac15

If k=\frac15, then

1=a(1+4)\implies a=\frac15

This means

\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}

Consider the nth partial sum of the series:

S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)

The sum telescopes so that

S_n=\dfrac2{14}-\dfrac2{5n+4}

and as n\to\infty, the second term vanishes and leaves us with

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17

7 0
3 years ago
Find y, at and tu I need help please help
lesantik [10]

Answer:

there is no picture... but i will try my best to help you if you post picture in another question

8 0
3 years ago
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Is 0.06 bigger than 1?
taurus [48]

Hi there!


Great question! 0.06 is actually SMALLER than 1.


Why?


Because 0.06 is almost half of the whole number. Half of 1 is 0.5, which is larger than 0.06.

 

Half is smaller than whole, so, therefore, 1 is larger.


Hope this helps!

Message me if you need anything else! I'd be happy to help! :D

8 0
4 years ago
Read 2 more answers
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