All categories

2 years ago

5 Kalimat tanyaJika Sedang Mewawancarai dokter Hewan

Mathematics

1 answer:

fiasKO [112]2 years ago

4 0

<h2>Pembahasan : </h2>

Kalimat tanya merupakan kalimat yang disampaikan untuk bermaksud mendapat jawaban tentang informasi, penjelasan, atau pernyataan.

<u>5 Kalimat tanya jika Sedang Mewawancarai dokter Hewan</u>

Apakah binatang peliharaan saya baik-baik saja?
Makanan sehat apa yang cocok untuk diberikan binatang peliharaan saya?
Kenapa binatang peliharaan saya jarang makan atau tidak mau makan ?
Bagaimana caranya agar menjaga kebersihan tubuh binatang peliharaan saya ?
Seberapa sering saya harus mengunjungi dokter untuk memeriksa hewan saya?

You might be interested in

Abby will receive a cash prize when she earns 300 bonus points at the videoshop. Abby receives 5 bonus points every time she ren

Agata [3.3K]

D. 48 she must rent 48 in order to receive the cash prize

6 0

3 years ago

What is the value of y when the value of x is 1 Car mileage for a hybrid car

Svetllana [295]

Answer: 60

Step-by-step explanation:

Just got the wrong answer to see what it was

8 0

3 years ago

ASAP<br> -will mark brianist

Amiraneli [1.4K]

Answer:

20-12

35-21

40-24

Step-by-step explanation:

You times both numbers by the same value

5 0

3 years ago

Read 2 more answers

I’m really bad at math word problems so yeah lol don’t make fun of me

Virty [35]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please help me answer this question

ratelena [41]

By the definition of the hyperbolic function tanh x, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

<h3>Hyperbolic functions & Proof of identities </h3>

By definition

$tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}$

Then,

$\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}$

$=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}$

$=\frac{2e^{-x}}{2e^{x}}$

$=\frac{e^{-x}}{e^{x}}$
$=e^{-x} \times \frac{1}{e^{x}}$

$= e^{-x} \times e^{-x}$

$= e^{-x+-x}$

$= e^{-x-x}$

$= e^{-2x}$

Hence, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

Learn more on Proof of Identities here: brainly.com/question/2561079

#SPJ1

6 0

2 years ago