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3 years ago

Can two different numbers have the same absolute value? If yes, give an example. If no, explain why not

2 answers:

7 0

If u wanna go by a technical term then yes, it can. The positive and negative side of a number. 4 example, -5. It's absolute value is 5. Positive 5's absolute value is 5 as well.

Now if u don't wanna be technical and don't want to use negative numbers, then no, it cannot.

IrinaK [193]3 years ago

3 0

Answer:

Yes, two different numbers have the same absolute value

Step-by-step explanation:

We are given that two different numbers

We have to tell that two different numbers can have same absolute value or not .

Absolute value of any number :

If a number x then its absolute value

$\mid x\mid$

Suppose we have -2 and 2

Absolute value of 2 is given by

$\mid 2\mid =2$

Absolute value of -2

$\mid {-2}\mid =2$

Hence, both numbers have same absolute value .

Then we can say that two numbers can have same absolute value.

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Rewrite in simplest rational exponent form √x • 4√x. Show each step of your process.

Studentka2010 [4]

I like to think of radicals (or square roots) just as x terms-
here, we have the equivalent (kind of) to x * 4x.
Since these are both x terms and 4x= x*x*x*x, we can simply write 5x.

Im not sure what your teacher is asking you to communicate, but most simply would like to see evidence that you understand the material rather than simply looking up the answer... (i mean without reading an explanation xD)
4√x= √x*√x*√x*√xso √x • 4√x =5√x

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3 years ago

Please Help! This is a trigonometry question.

liraira [26]

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$\large\begin{array}{l} \textsf{Square both sides of \mathsf{(i)} above:}\\\\ \mathsf{(3\,sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{9\,sin^2\,\theta=4\,cos^2\,\theta}\qquad\quad\textsf{(but }\mathsf{cos^2\theta=1-sin^2\,\theta}\textsf{)}\\\\ \mathsf{9\,sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{9\,sin^2\,\theta=4-4\,sin^2\,\theta}\\\\ \mathsf{9\,sin^2\,\theta+4\,sin^2\,\theta=4} \end{array}$

$\large\begin{array}{l} \mathsf{13\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{13}}\\\\ \mathsf{sin\,\theta=\sqrt{\dfrac{4}{13}}}\\\\ \textsf{(we must take the positive square root, because }\theta \textsf{ is an}\\\textsf{acute angle, so its sine is positive)}\\\\ \mathsf{sin\,\theta=\dfrac{2}{\sqrt{13}}} \end{array}$

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$\large\begin{array}{l} \textsf{From (i), we find the value of }\mathsf{cos\,\theta:}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{2}\,sin\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{13}}}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\sqrt{13}}}\\\\ \end{array}$

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$\large\begin{array}{l} \textsf{Since sine and cosecant functions are reciprocal, we have}\\\\ \mathsf{sin\,2\theta\cdot csc\,2\theta=1}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{sin\,2\theta}\qquad\quad\textsf{(but }}\mathsf{sin\,2\theta=2\,sin\,\theta\,cos\,\theta}\textsf{)}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\,sin\,\theta\,cos\,\theta}}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\cdot \frac{2}{\sqrt{13}}\cdot \frac{3}{\sqrt{13}}}} \end{array}$

$\large\begin{array}{l} \mathsf{csc\,2\theta=\dfrac{~~~~1~~~~}{\frac{2\cdot 2\cdot 3}{(\sqrt{13})^2}}}\\\\ \mathsf{csc\,2\theta=\dfrac{~~1~~}{\frac{12}{13}}}\\\\ \boxed{\begin{array}{c}\mathsf{csc\,2\theta=\dfrac{13}{12}} \end{array}}\qquad\checkmark \end{array}$

<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2150237

$\large\textsf{I hope it helps.}$

Tags: <em>trigonometry trig function cosecant csc double angle identity geometry</em>

</span>

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Answer:

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Step-by-step explanation:

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