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nika2105 [10]
3 years ago
7

How many feet can a sloth go in 90 minutes?

Mathematics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

600

Step-by-step explanation:

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May someone help me with 2-9? Please?
mylen [45]

2. Each side of a pentagon is the same size.

4cm x 5 = 20cm or 4cm+4cm+4cm+4cm+4cm = 20cm

3. Each side of a square is the same size.

13yd x 4 = 52yd or 13yd+13yd+13yd+13yd = 52yd

4. Add all sides together.

12m+12m+30m+30m = 84m

5. Again add all sides together.

16yd+16yd+4yd+4yd = 40yd

6. Each side of a square is the same size.

7in x 4 = 28in. or 7in+7in+7in+7in = 28in

7. Add all sides together.

2cm+2cm+3cm+3cm = 10cm

8. Each side of a rhombus is the same size. A rhombus has 4 sides.

23in x 4 = 92in or 23in+23in+23in+23in = 92in

9. A regular octagon has 8 sides and each side is the same size.

9cm x 8 = 72cm

7 0
4 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
It is -40°F on a cold night in the North Pole. When the sun is out the next day, it is -5°F. What is the difference in temperatu
Alexeev081 [22]

Answer:

-35°F

Step-by-step explanation:

-40 - -5 equals -35°F

8 0
3 years ago
Please help me pretty please
antoniya [11.8K]

Answer:

28.6

Step-by-step explanation:

Pathagorem therom

7 0
3 years ago
prove the following identity: sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x please provide a proof in some shape form or fash
wolverine [178]

Answer:

Step-by-step explanation:

Hello,

<u><em>Is this equality true ?</em></u>

sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x

<u>1. let 's estimate the left part of the equation</u>

sec(x)csc(x)(tan(x) + cot(x)) =\dfrac{1}{cos(x)sin(x)}*(\dfrac{sin(x)}{cos(x)}+\dfrac{cos(x)}{sin(x)})\\\\=\dfrac{1}{cos(x)sin(x)}*(\dfrac{sin^2(x)+cos^2(x)}{sin(x)cos(x)})\\\\=\dfrac{1}{cos(x)sin(x)}*(\dfrac{1}{sin(x)cos(x)})\\\\\\=\dfrac{1}{cos^2(x)sin^2(x)}

<u>1. let 's estimate the right part of the equation</u>

<u />2+tan^2(x) + cot^2(x)=2+\dfrac{sin^2(x)}{cos^2(x)}+\dfrac{cos^2(x)}{sin^2(x)}\\\\=\dfrac{2cos^2(x)sin^2(x)+cos^4(x)+sin^4(x)}{cos^2(x)sin^2(x)}\\\\=\dfrac{(cos^2(x)+sin^2(x))^2}{cos^2(x)sin^2(x)}\\\\=\dfrac{1^2}{cos^2(x)sin^2(x)}\\\\=\dfrac{1}{cos^2(x)sin^2(x)}<u />

This is the same expression

So

sec x csc x(tan x + cot x) = 2+tan^2 x + cot^2 x

hope this helps

7 0
3 years ago
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