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3 years ago

11

Solve 2(3x+3)=2(4x) and 2(7x-2)=2x+4

2 answers:

Agata [3.3K]3 years ago

7 0

For the first equation x=3. For the second equation x=i dont know

VARVARA [1.3K]3 years ago

6 0

6x+6=8x
-6x -6x
-----
6=2x
6 divided by 2
x=3

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What is the volume of the right rectangular prism?

kenny6666 [7]

Answer:

240

Step-by-step explanation:

V=width height length = 10 · 3 · 8 = 240

Hope this helped, have a wonderful day man!

4 0

2 years ago

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Which values are solutions to the inequality below ×<9

iragen [17]

Answer: theres not a picture soo idk

Step-by-step explanation:

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3 years ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

Need help on this... First one to answer this correctly, I will give you a BRANLIST

Simora [160]

Answer:

wheres the problem?

Step-by-step explanation:

6 0

3 years ago

Question part points submissions used use newton's method with the specified initial approximation x1 to find x3, the third appr

serious [3.7K]

Set $f(x)=2x^3-3x^2+2$ . Find the tangent line $\ell_1(x)$ to $f(x)$ at the point when $x=x_1$ :

$f'(x)=6x^2-6x\implies f'(x_1)=12$ (slope of $\ell_1$ )

$\implies\ell_1(x)=12(x-x_1)+f(x_1)=12(x+1)-3=12x+9$

Set $x_2=-\dfrac9{12}$ , the root of $\ell_1(x)$ . The tangent line $\ell_2(x)$ to $f(x)$ at $x=x_2$ has slope and thus equation

$f'(x_2)=\dfrac{63}8\implies\ell_2(x)=\dfrac{63}8\left(x+\dfrac9{12}\right)-\dfrac{17}{32}=7x+\dfrac{151}{32}$

which has its root at $x_3=-\dfrac{151}{224}\approx-0.6741$ .

(The actual value of this root is about -0.6777)

5 0

2 years ago