why does a temperate zone support many varieties of organisms

Answers

Answer 1

Answer:

Torrid zone is too hot which sometimes causes more hotness in soil and won't have water which helps the plants and organisms not to survive. => In such cases Temperate zone experience both climatic conditions for short periods of time, Hence it supports many varieties of organisms


Related Questions

Greenhouse Gases Worksheet
help me pls

Answers

Answer:

1.)second page

1. CO2

2. Million

3. increase

4. human activities

5. deforestation

6. fuels

7. oil

8. oxygen

9. deforestation

10. increase

11. 2050

2.) sorry I'm out of time I can't answer this part sorry :( this is page 2

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

3.)

1.chemicals

2.Radiation

3.increasing

4.trapping

Difference between cultural and natural attractions in tourism

Answers

natural attractions is the place and things that were naturally there and cultural attractions is what the humans contribute to create their civilization

what is the largest fresh water volcanic island in the world? najuato ometepe managua colombian

Answers

Answer:

Ometepe

Explanation:

It's 276 sq km large

Answer:

Ometepe

Explanation:

Got it right on Edge :)

CAN SOMEONE PLEASE HELP I WILL GIVE 30 POINTS AND MARK BRAINLIEST IF THE ANSWER IS CORRECT ALSO NO LINKS PLEASE

Answers

Answer:

I would guess Economic because of mining and mineral deposits

Who were the Mexica?

Answers

Answer:

here

Explanation:

The Mexica were a Nahua people who founded Tenochtitlan and Tlatelolco in lake texcoco in 1325 and 1337.

Find an equation for the perpendicular bisector of the line segment whose endpoints
are (5, -4) and (-9, -8)

Answers

Answer:

y= -7/2x-26/2 or y= -7/2x-13

Explanation:

To solve this question you need all of these formulas:

gradient: [tex]\frac{y2-y1}{x2-x1}[/tex] , Point-Slope [tex]y-y1=m(x-x1)[/tex] ,

perpendicular m [tex](m1)x(m2)=-1[/tex] , Midpoint [tex](\frac{x1+x2}{2}[/tex],[tex]\frac{y1+y2}{2} )[/tex]

and equation of a line [tex]y=mx+c[/tex]

where m stands for gradient

First things first.

To start you have to know the data of the line that includes the two endpoints, so you calculate its gradient (m) of this line with the gradient fromula: [tex]m=\frac{-4-(-8)}{5-(-9)}[/tex] , which equals 4/14 or 2/7

(It can also be called rise/run) (Remember the rule of signs where - and - equal +)

with that information you can proceed with the point-slope or point-gradient formula, so you plug the values: y-(-8) = 2/7 (x-(-9)), which results in y+8=2/7(x+9) and then y+8=2/7x+18/7.

To finish the equation you move eight to the other side. To simplify things you can change it into a fraction as I did, and remember to change signs.

y=2/7x+18/7 -8 -> y=2/7x+18/7-56/7 . This gives us the number of y=2/7x-38/7, which is the equation of the first line.

Now to know the gradient of the second line you apply the formula of perpendicular bisector where m1 x m2 = -1. We know m1 (gradient of the first line) is 2/7, so m2 = [tex]\frac{-1}{2/7}[/tex] = -7/2. m2 is therefore -7/2

Now you have to know the midpoint between the two endpoints, which will act as the start point of the perpendicular bisector

M (midpoint) = [tex](\frac{-9+5}{2} ,[/tex][tex]\frac{8+(-4)}{2} )[/tex], which give us the coordinates of (-2, -6)

(remember, x coordinate is always first)

with this point we can apply again the point-slope formula to know the equation of the line:

y-(-6)=-7/2(x-(-2)) -> y+6=-7/2(x+2) -> y+6=-7/2x - 14/2

Move the 6 to isolate the y

y=-7/2x -14/2 - 12/2

which equals y= -7/2x -26/2

You can check the results in this page: GraphPlotter

https://www.transum.org/Maths/Activity/Graph/Desmos.asp

To make sure the answer is correct.

Hope it helps :)

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