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melomori [17]
4 years ago
6

If f^n(0)=(n+1)! maclaurin series

Mathematics
1 answer:
guajiro [1.7K]4 years ago
7 0
Pasting my answer from the linked question in case that one gets deleted for some reason

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tom jogged 3/10 mile, betsy jogged 5/10 mile, and sue jogged 2/10 mile. who jogged a longer distance than 4/10 mile?
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Betsey jogged the longer distance.
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4 years ago
Please someone help, give the right answer it’s important
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Answer:

(2)

Step-by-step explanation:

Having three of the same angles ("AAA") does not prove that two triangles are congruent. All the other options use one of the viable methods (SSS, SAS, ASA, AAS).

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Find the missing side lengths. Leave your answers as radicals in simplest form.
Grace [21]

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3 years ago
Can someone please help with these 2 word problems ASAP.
PolarNik [594]

Answer:

4.     4.66hours

5.  time_trip = 2.777 h

Step-by-step explanation:

4. A jet plane travels 2 times the speed of a commercial airplane. The distance between Vancouver and Regina is 1730 km. If the flight from Vancouver to Regina on a commercial airplane takes 140 minutes longer than a jet plane, what is the time of a commercial plane ride of this route?

We know that

Speed_commercial = distance/ time_comm

Speed_commercial = 1730 km/ time_comm

Then,

Speed_jet = distance/ time_jet = 2*Speed_commercial

Speed_jet =   1730 km/ time_jet = 2*Speed_commercial

time_comm = 2.333 h + time_jet

1730 km/ time_jet = 2*1730 km/ time_comm

time_comm = 2*time_jet

time_comm = 2*(time_comm  - 2.333h)

time_comm = (2*time_comm  - 4.666h)

time_comm  = 4.666 h

5. A man goes fishing in a river and wants to know how long it will take him to get 10km upstream to his favourite fishing location. The speed of the current is 3 km/hr and it takes his boat twice as long to go 3km upstream as is does to go 4km downstream. How long will it take his boat to get to his fishing spot?

Distance = 10 km upstream

Speed_current = 3 km/h

Upstream

(Speed_boat - Speed_current)  = 3 km / (2*time_downstream)

Downstream

(Speed_boat + Speed_current)  = 4 km / (time_downstream)

We subtract both equations

(Speed_boat + Speed_current) - (Speed_boat - Speed_current) = 4 km / (time_downstream) - 3 km / (2*time_downstream)

2*Speed_current = (5/2 km)/ time_downstream

2*(3 km/h)= (5/2 km)/ time_downstream

time_downstream = 0.416 h

(Speed_boat + 3km/h)  = 4 km / (0.416 h)

Speed_boat  = 6.6 km/h

Trip upstream

(Speed_boat - Speed_current)  = 10 km / (time_trip)

time_trip = 10 km/(3.6 km/h) = 2.777 h

time_trip = 2.777 h

3 0
3 years ago
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