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4 years ago

13

2 (2x+1)+2x+x2+x simplified

1 answer:

Maurinko [17]4 years ago

8 0

X^2 + 7x + 2 that’s your answer

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Find all solutions in the interval [0, 2π). <br> 2 sin2x = sin x

tester [92]

Answer:

The solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

Step-by-step explanation:

We will need the double angle identity $\sin(2x)=2\sin(x)\cos(x)$ .

Let's begin:

$2\sin(2x)=\sin(x)$

Use double angle identity mentioned on left hand side:

$2\cdot 2\sin(x)\cos(x)=\sin(x)$

Simplify a little bit on left side:

$4\sin(x)\cos(x)=\sin(x)$

Subtract $\sin(x)$ on both sides:

$4\sin(x)\cos(x)-\sin(x)=0$

Factor left hand side:

$\sin(x)[4\cos(x)-1]=0$

Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:

$\sin(x)=0 \text{ or } 4\cos(x)-1=0$

The first is easy what angles $\theta$ are $y$ -coordinates on the unit circle 0. That happens at $0$ and $\pi$ on the given range of $x$ (this $x$ is not be confused with the $x$ -coordinate).

Now let's look at the second equation:

$4\cos(x)-1=0$

Isolate $\cos(x)$ .

Add 1 on both sides:

$4 \cos(x)=1$

Divide both sides by 4:

$\cos(x)=\frac{1}{4}$

This is not as easy as finding on the unit circle.

We know $\arccos( )$ will render us a value between $0$ and $2\pi$ .

So one solution on the given interval for x is $x=\cos^{-1}(\frac{1}{4})$ .

We know cosine function is even.

So an equivalent equation is:

$\cos(-x)=\frac{1}{4}$

Apply $\cos^{-1}$ to both sides:

$-x=\cos^{-1}(\frac{1}{4})$

Multiply both sides by -1:

$x=-\cos^{-1}(\frac{1}{4})$

This going to be negative in the 4th quadrant but if we wrap around the unit circle, $2\pi$ , we will get an answer between $0$ and $2\pi$ .

So the solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

5 0

3 years ago

If QR=13 and PT=8, FIND QT of the rhombus

evablogger [386]

Answer:

$|QT|=10.2$ units.

Step-by-step explanation:

The adjacent sides of a kite are equal.

$\Rightarrow |PQ|=13=|QR|$

Triangle PQT is a right triangle.

From the Pythagoras Theorem;

$|QT|^2+|PT|^2=|PQ|^2$

$|QT|^2+8^2=13^2$

$|QT|^2+64=169$

$|QT|^2=169-64$

$|QT|=\sqrt{104}$

$|QT|=2\sqrt{26}$

$|QT|=10.2$ units.

5 0

4 years ago

Write an solve an equation based off the verbal phrase. 10 less than half a number is 27.

olga_2 [115]

Answer: <em>x/2 - 10 = 27, and x is the number. </em>

I think that's right. :)

8 0

3 years ago

Read 2 more answers

The table shows input and output values of the function y = x2 + 12x – 2. What is an approximate solution of the equation x2 + 1

romanna [79]

Answer:

The solutions for the equations are x = 0.1644 and x = -12.1644

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$x^{2} + 12x - 2 = 0$

So

$a = 1, b = 12, c = -2$

$\bigtriangleup = 12^{2} - 4*1*(-2) = 152$

$x_{1} = \frac{-12 + \sqrt{152}}{2} = 0.1644$

$x_{2} = \frac{-12 - \sqrt{152}}{2} = -12.1644$

The solutions for the equations are x = 0.1644 and x = -12.1644

5 0

3 years ago

Find the slope that passes through the points (-18, -6) and (7, -17). *

kotykmax [81]

B.-11/25 is the answer

8 0

3 years ago