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Hunter-Best [27]
3 years ago
15

Why cant i see answers anymore

Mathematics
2 answers:
Eddi Din [679]3 years ago
7 0

Answer:

reload the page it happens to me but that might mean your gonna have to watch the add again wait nvm it aint working

Step-by-step explanation:

liraira [26]3 years ago
7 0

Answer:

having the same prob rn

Step-by-step explanation:

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Find f(x)-g(x)<br> f(x)=7x+15 g(x)=5x-6
-BARSIC- [3]
<h3>Answer: 2x+21</h3>

=============================

Work Shown:

f(x) - g(x) = [ f(x) ] - [ g(x) ]

f(x) - g(x) = [ 7x+15 ] - [ 5x-6 ]

f(x) - g(x) = 7x+15 - 5x+6

f(x) - g(x) = (7x-5x)+(15+6)

f(x) - g(x) = 2x+21

4 0
3 years ago
Leticia simplified an expression. Her work is shown below.
katen-ka-za [31]
Step 2, she didn't multiply 8 by 0.75.

Hope this helps :)
6 0
3 years ago
Read 2 more answers
Pls pls help, brainiest guaranteed
andrey2020 [161]

Answer:

i think its

4× + 5y > = 85

Step-by-step explanation:

i hope im right or even close i tryed i should be right tho HAVE A NICE DAY!!

3 0
3 years ago
Mr. Spence asks his math students to identify TWO shapes based on the
ale4655 [162]

Answer:

Step-by-step explanation:

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6 0
3 years ago
Determine if diverges, converges, or converges conditionally.
TEA [102]

The given series is conditionally convergent. This can be obtained by using alternating series test first and then comparing the series to the harmonic series.

<h3>Determine if diverges, converges, or converges conditionally:</h3>

Initially we need to know what Absolute convergence and Conditional convergence,

If \sum|a_{n} | → converges, and \sum a_{n} → converges, then the series is Absolute convergence

If \sum|a_{n} | → diverges, and \sum a_{n} → converges, then the series is Conditional convergence

First use alternating series test,

\lim_{k \to \infty} \frac{k^{5} +1}{k^{6}+11 } = \lim_{n \to \infty} \frac{5}{6k} = 0,

The series is a positive, decreasing sequence that converges to 0.

Next by comparing the series to harmonic series,

\sum^{\infty} _{k=2}|(-1)^{k+1} \frac{k^{5} +1}{k^{6}+11 }|=\sum^{\infty} _{k=2}\frac{k^{5} +1}{k^{6}+11 } ≈  \sum^{\infty} _{k=2}\frac{1}{k} = 0

This implies that the series is divergent by comparison to the harmonic series.

First we got that the series is converging and then we got the series is divergent. Therefore the series is conditionally convergent.

\sum|a_{n} | → diverges, and \sum a_{n} → converges, then the series is Conditional convergence.

Hence the given series is conditionally convergent.

Learn more about conditionally convergent here:

brainly.com/question/1580821

#SPJ1

7 0
2 years ago
Read 2 more answers
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