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2 years ago

What is the value of x? Enter your answer in the box. x =

Mathematics

1 answer:

qaws [65]2 years ago

7 0

Answer:

Step-by-step explanation: its yo mama :)

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Natali [406]

Answer:

D brainliest?

Step-by-step explanation:

6 0

3 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

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

The marked are the trigonometric identities

3 0

1 year ago

What is the answer to question 41

Nina [5.8K]

Answer:

an = 6 +5(n-1)

an = 1 +5n

Step-by-step explanation:

The equation for an arithmetic term is

an =a1 +d(n-1)

where a1 is the 1st term and d is the common difference

an = 6 +5(n-1)

We can distribute the 5

an = 6+5n-5

an = 1 +5n

4 0

3 years ago

Please help me answer this question

avanturin [10]

By direct substitution and simplification, the trigonometric function z = cos (2 · x + 3 · y) represents a solution of the partial differential equation $\frac{\partial^{2} t}{\partial x^{2}} - \frac{\partial^{2} t}{\partial y^{2}} = 5\cdot z$ .

<h3>How to analyze a differential equation</h3>

Differential equations are expressions that involve derivatives. In this question we must prove that a given expression is a solution of a differential equation, that is, substituting the variables and see if the equivalence is conserved.

If we know that $z = \cos (2\cdot x + 3\cdot y)$ and $\frac{\partial^{2} t}{\partial x^{2}} - \frac{\partial^{2} t}{\partial y^{2}} = 5\cdot z$ , then we conclude that: