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3 years ago

What is faster the 80 km/h or the 60 mi/h

Mathematics

1 answer:

nikdorinn [45]3 years ago

3 0

60 mph is faster. 80 kph is rounded to 49.7097 mph.

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Help, I’m no good with slope

postnew [5]

All you have to do is graph whatever those #'s are, so whatever numbers are top are the number that are on the bottom of the graph, and whatever #'s are on top are the numbers going up on the graph.

5 0

4 years ago

10. Select all true statements.

mart [117]

Answer:

its B and E

edit: i forgot C

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Richardo had saved $80 from his summer job to buy video games. He

Vlad1618 [11]

Answer:

$15.90

Step-by-step explanation:

80 - 16.40 = 63.60

63.60÷4 =15.9

15.9 = $15.90

7 0

3 years ago

How does the graph of g(x) = (x + 4)3 − 6 compare to the parent function f(x) = x3?

Solnce55 [7]

Answer:

g(x) is shifted 6 units to the left

Step-by-step explanation:

Lets try to simplify g(x) since has a few extra terms:

g(x)= 3x+12-6=3x+6

Now it is easier to compare the two functions.

We can tell that they both have the same slope, both differs on a extra term

This term tell us that the g(x) is shifted to the left (it is positive 6)

Another approach to the solution is to plot the two functions together by obtaining the crossing points with the 'y' axis and with the 'x' axis

the result is shown in the attached picture

4 0

4 years ago

Read 2 more answers

Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.

Arisa [49]

Answer: $S_n=5(1-\dfrac{1}{n+1})$ ; 5

Step-by-step explanation:

Given series : $[\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}]$

Sum of series = $S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}]$

Consider $\dfrac{1}{n\cdot(n+1)}=\dfrac{n+1-n}{n(n+1)}$

$=\dfrac{1}{n}-\dfrac{1}{n+1}$

⇒ $S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}]$

Put values of n= 1,2,3,4,5,.....n

⇒ $S_n=5(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......-\dfrac{1}{n}+\dfrac{1}{n}-\dfrac{1}{n+1})$

All terms get cancel but First and last terms left behind.

⇒ $S_n=5(1-\dfrac{1}{n+1})$

Formula for the nth partial sum of the series :

$S_n=5(1-\dfrac{1}{n+1})$

Also, $\lim_{n \to \infty} S_n = 5(1-\dfrac{1}{n+1})$

$=5(1-\dfrac{1}{\infty})\\\\=5(1-0)=5$

4 0

3 years ago