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3 years ago

What is the graph of y=arctan(x)

Mathematics

2 answers:

irina1246 [14]3 years ago

7 0

Answer:

The answer is A on edg2020

Step-by-step explanation:

It is the first graph. Hope this helps! #classof2020

Marat540 [252]3 years ago

4 0

Answer:

All real numbers (D)

Step-by-step explanation:

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Find the measure of angle b (image below)

IRINA_888 [86]

Angle b will be 34 degrees because angle b and the other angle that is 34 degrees are vertical angles!

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3 years ago

Find the number between 30 and 70 that is divisible by 8, and when divided by 5 has a remainder of 1

monitta

30+70÷8÷5-1=
=30.75
hope this helps

7 0

2 years ago

Could someone help me, please?

k0ka [10]

We can rewrite the expression under the radical as

$\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4$

then taking the fourth root, we get

$\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|$

Why the absolute value? It's for the same reason that

$\sqrt{x^2}=|x|$

since both $(-x)^2$ and $x^2$ return the same number $x^2$ , and $|x|$ captures both possibilities. From here, we have

The absolute values disappear on all but the $b$ term because all of $\dfrac32$ , $a^2$ and $c^4$ are positive, while $b^3$ could potentially be negative. So we end up with

$\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4$

3 0

3 years ago

Anybody know this plz help me

aniked [119]

y= -3/1x + 5

i believe you just miscounted

4 0

3 years ago

Read 2 more answers

Power (denoted by PPP) can be defined as a function of work (denoted by WWW) and time (denoted by ttt) using this formula: P=\df

Amanda [17]

Answer:

$\dfrac{\text{kg}\cdot\text{m}^2}{\text{s}^3}$

Step-by-step explanation:

Power (denoted by P) can be defined as a function of work (denoted by W) and time (denoted by t) using this formula:

$P=\dfrac{W}{t}$

Unit of Work = $\dfrac{\text{kg}\cdot\text{m}^2}{\text{s}^2}$

Unit for Time = s

Therefore, the Unit of Power

$=$Unit for W \times \dfrac{1}{\text{Unit for time}}$

$=\dfrac{\text{kg}\cdot\text{m}^2}{\text{s}^2} \times \dfrac{1}{s} \\=\dfrac{\text{kg}\cdot\text{m}^2}{\text{s}^3}$

An appropriate unit for power is: $\dfrac{\text{kg}\cdot\text{m}^2}{\text{s}^3}$

3 0

2 years ago