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3 years ago

How are a desert and a tundra similar?

Mathematics

1 answer:

Hatshy [7]3 years ago

8 0

Answer:

D is right option

Step-by-step explanation:

Similarities:

<em>Both the biomes experience less precipitation due to this they have a less diversity of flora and fauna as compared to other biomes like savanna, grasslands, chaparral etc. Let us see how they differ from each other!</em>

Low rain fall _ annual rainfal of lessthan 20cm in deserts, 15-20cm in tundra

Minimal life_ only small shurbs can survive in both kind of environments

Poor drainage.. In deserts, though sandy soil may seem much porous, these are most flood prone areas of world. The tundra is characterised by permafrost soils ie, under the thin layer of soil, a thick ice sheet is present and hence is no seepage.

High range of temperatures: In tundra, the maximum temperatures are recorded in summer (abt 10^C) and minimum during winter(-20 to -30^C). While, in desert too, the temperature range is high but diurnally ie, nights are too cold and day time is scorching.

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I know the selected answer is correct but I'm not too sure how to get that answer.

Kryger [21]

$\tt{ Hey \: there , \: Mr.Panda \: ! }$ ;)

♨ $\large{ \tt{ E \: X \: P \: L \: A \: N \: A \: T \: I\: O \: N}}:$

⤻ Before solving the given question , you should know the answer of these questions :

✺How do you find the hypotenuse , perpendicular and base when the angle ( $\theta \: , \alpha \: ,\beta$ ) is given ?

⇾ The longest side , which is the opposite side of right angle is the hypotenuse ( h ). There are two other sides , the opposite and the adjacent. The naming of these sides depends upon which angle is involved. The opposite is the side opposite the angle involved and it is called the perpendicular ( p ) . The adjacent us the side next to the angle involved ( buy not the hypotenuse ) and it is called the base ( b ).

☄ $\large{ \tt{REMEMBER}} :$

$\bf{ \sin \theta = \frac{opposite}{hypotenuse} = \frac{perpendicular}{hypotenuse} }$

$\bf{ \cos\theta = \frac{adjacent}{hypotenuse} = \frac{base}{hypotenuse} }$

$\bf{ \tan \theta = \frac{opposite}{adjacent} = \frac{perpendicular}{base} }$

In the above cases , $\theta$ is taken as the angle of reference.

♪ Our Q/A part ends up here! Let's start solving the question :

❈ $\large{ \tt{GIVEN}} :$

Perpendicular ( p ) = ? , Hypotenuse ( h ) = 18 & base ( b ) = 16

✧ $\large{ \tt{TO \: FIND} : }$

Value of tan $\theta$

✎ $\large{ \tt{SOLUTION}} :$

Firstly , Finding the value of perpendicular ( p ) using Pythagoras theorem :

❃ $\boxed{ \sf{ {h}^{2} = {p}^{2} + {b}^{2} }}$ [ Pythagoras theorem ]

$\large{ ⇢ \sf{p}^{2} + {b}^{2} = {h}^{2} }$

$\large{⇢ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}$

$\large{ ⇢\sf{ {p}^{2} = {18}^{2} - {16}^{2} }}$

$\large{⇢ \sf{ {p}^{2} = 324 - 256}}$

$\large{⇢ \sf{ {p}^{2} = 68}}$

$\large{⇢ \sf{p = \sqrt{68}}}$

$\large{ ⇢\sf{p = \boxed{ \tt{2 \sqrt{17}}} }}$

Okey, We found out the perpendicular i.e $\tt{2 \sqrt{17}}$ . Now , We know :

❊ $\large{ \sf{ \tan \theta} = \frac{perpendicular}{base} }$

$\large {\tt{↬ \: tan \theta = \frac{2 \sqrt{17} }{16}}}$

$\large{ \tt{ ↬ tan \theta = \frac{ \cancel{2} \: \sqrt{17} }{ \cancel{16} \: \: 8} }}$

$\large{ \tt{ ↬ \boxed{ \tt{tan \theta = \frac{ \sqrt{17} }{8}}}}}$

⟿ $\boxed{ \boxed{ \tt{OUR\: FINAL \: ANSWER : \boxed{ \underline{ \bf{ \frac{ \sqrt{17} }{8}}}}}}}$

۵ Yay! We're done!

♕ $\large\tt{RULE \: OF \:SUCCESS }:$

Never lose hope & keep on working ! ✔

ツ Hope I helped!

☃ Have a wonderful day / evening! ☼

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2 years ago

JK = 8x – 5 and<br> KL = 7x + 3<br> Find JL

vladimir2022 [97]

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6 0

3 years ago

How do I put 3x - 6y = -12 into Slope Intercept Form?

max2010maxim [7]

Answer:

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Step-by-step explanation:

hope this helps

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2 years ago

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4x2 + 4x - 1, evaluate and fully simplify each of the - For the function f(0) following f(s + h) f(x + h) - f(5) h 10

Andrew [12]

Given the function f(x);

$f(x)=-4x^2+4x-1$

Evaluating the function f(x+h);

$\begin{gathered} f(x+h)=-4(x+h)^2+4(x+h)-1 \\ f(x+h)=-4(x^2+2xh+h^2)^{}+4(x+h)-1 \\ f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1 \end{gathered}$

So;

$f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1$

Evaluating the second function;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1-(-4x^2+4x-1)}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1+4x^2-4x+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2+4x^2-4h^2-8xh^{}+4x-4x+4h-1+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h} \end{gathered}$

simplifying further;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h}=-4h-8x+4 \\ \frac{f(x+h)-f(x)}{h}=-4h-8x+4 \end{gathered}$

Therefore, we have;

$undefined$

7 0

1 year ago