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3 years ago

10

A desest tortoise spends about 95% of its life in underground burrows.About what fraction of a desert tortoise's life is spent a

bove ground

1 answer:

8090 [49]3 years ago

3 0

5/100's tortoise's life was spent above ground.

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What all numbers whose absolute value is 6

GalinKa [24]

Answer:

6, -6

Explanation:

-----

3 0

3 years ago

Mustafa buys a book that costs $12.50 if the sales tax is 8% what is the total cost of the book

wariber [46]

The total cost of the book would be:

12.50* 0.8= 10

The total cost of the book would be $10.00

7 0

3 years ago

Read 2 more answers

I need help asap pls and thank you ;)

olga55 [171]

Answer:

$\text{Length of AB is }\frac{ah}{a+h}$

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP (∵ corresponding angles)

∠NDC=∠NPM (∵ corresponding angles)

By AA similarity rule, ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

$\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\$

$KA(\frac{h}{x}-1)=AP$

$ND(\frac{h}{x}-1)=PD$

Adding above two, we get

$(KA+ND)(\frac{h}{x}-1)=(AP+PD)$

⇒ $(KN-AD)=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x^2}{h-x}$

⇒ $x^2=ah-ax-xh+x^2$

⇒ $x(h+a)=ah$

⇒ $x=\frac{ah}{a+h}$

3 0

2 years ago

Select all sets that contain the number 0.484848.

Yuki888 [10]

0.484848 . . . . .

<span>A. Integer . . . no, it's not
B. Irrational . . . no, it's not
C. Natural. . . no, it's not
D. Rational . . . Yes ! It is !
E. Whole</span> . . . no, it's not

4 0

3 years ago

Inverse laplace of [(1/s^2)-(48/s^5)]

Katen [24]

**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

Hope this helps...

7 0

3 years ago