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3 years ago

What is the answer to m+4=-12

1 answer:

8 0

Hey there!

You need to get m by itself. So subtract 4 on each side. You get m=-16. (-12-4 is the same as -12+-4). Since m is by itself, this is the answer: m=-16.

I hope this helps!

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<img src="https://tex.z-dn.net/?f=x%20%2B%20y%20%20%3D%2014%5C%5C%202x%20%2B%204y%20%3D%2018" id="TexFormula1" title="x + y = 1

alexgriva [62]

Answer:

x = 19, y = -5

Step-by-step explanation:

x + y = 14 => x= 14-y

2x + 4y = 18 => 2(x+2y)=18 => x+2y=9

x+2y=9 => (14 - y) + 2y = 9

14 + y = 9 => y = -5

x + (-5) = 14 => x=19

5 0

3 years ago

Which of the following pairs of functions are inverses of each other

Ghella [55]

Option (A) represents the function and its inverse of a function option (A) is correct.

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

We have a function and inverse of a function shown in the picture.

Checking for option (A):

$\rm f(x) = \dfrac{x}{4}+10 \ \ \ and \ \ \ g(x) = 4x -10$

Taking f(x):

$\rm f(x) = \dfrac{x}{4}+10$

To find the inverse of a function interchange the x and y variables:

f(x) → x

x → g(x)

$\rm x = \dfrac{g(x)}{4}+10$

Solving:

4x = g(x) + 10

g(x) = 4x - 10

Similarly, we can find the inverse of a function.

Thus, option (A) represents the function and its inverse of a function option (A) is correct.

Learn more about the function here:

brainly.com/question/5245372

#SPJ1

8 0

2 years ago

Read 2 more answers

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities