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3 years ago

PPlz help meeee need answer

Mathematics

1 answer:

user100 [1]3 years ago

5 0

P^2 could also have -4 as an answer. Any time a number is squared, it is positive

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Evaluate tan ( cos^-1 ( 1/8 ) ) , giving your answer as an exact value (no decimals)

vekshin1

$\bf cos^{-1}\left( \cfrac{1}{8} \right)=\theta \qquad \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{8}}\impliedby \textit{let's find the \underline{opposite}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{8^2-1^1}=b\implies \pm\sqrt{63}=b ~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\pm\sqrt{63}}}{\stackrel{adjacent}{1}} \\\\\\ ~\hspace{34em}$

$\bf \stackrel{\textit{keeping in mind that}}{tan\left(cos^{-1}\left( \frac{1}{8} \right) \right)}\implies tan(\theta )$

5 0

2 years ago

Max points and brainliest

Paul [167]

Answer:

-2

Step-by-step explanation:

slope = (y1-y2)/(x1-x2)

= (3-1)/(0-1)

= 2/-1

= -2

3 0

3 years ago

Read 2 more answers

Expand the following by using distributive property 6(-3w+1/3)

netineya [11]

Answer:

-18w+2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

PLEASE HELP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Bond [772]

Answer:

$R = \frac {3ST}{2S + T}$

Step-by-step explanation:

Given the equation;

$\frac {3}{R} - \frac {1}{S} = \frac {2}{T}$

Rearranging the equation, we have;

$\frac {3}{R} = \frac {2}{T} + \frac {1}{S}$

Lowest common multiple (LCM) of S and T is ST.

$\frac {3}{R} = \frac {2S + T}{ST}$

Cross-multiplying, we have;

$3ST = R(2S + T)$

Making R, the subject of formula;

$R = \frac {3ST}{2S + T}$

8 0

3 years ago

I need #14 done by tomorrow around 9-10:00. Thank you so much if you do. God bless you :3

Vinvika [58]

Answer:

A) $\frac{1}{5}$

B) - 5

C) Not Possible

D) 5

E) $\frac{-1}{5}$

Step-by-step explanation:

All integers are rational numbers. But not all rational numbers are integers.

All whole numbers are integers. But not all integers are whole numbers.

I am a rational number but not an integer. Located on the right of 0.

This means that it should be a positive number. Since, it is a rational number but not an integer, it should be of the form $\frac{p}{q}$ .

From, the options $\frac{1}{5}$ would fit this description.

I am a rational number and an integer but not a whole number.

This means that it should be a negative integer. Since, all positive integers and zero would be whole numbers. From the options, the answer would be -5.

I am a whole number but not an integer.

This is clearly not possible because all whole numbers are a subset of integers.

I am a rational number, a whole number and an integer.

This means it is a positive integer. 5 would fit this description.

I am a rational number but not an integer; located on the left side of 0.

This means it is a negative number. $-\frac{1}{5}$ should be the answer.

3 0

2 years ago