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2 years ago

Can you put your tongue in a not??

Mathematics

1 answer:

Masteriza [31]2 years ago

3 0

Answer:

No, it is physically impossible to put your tongue in a knot.

Step-by-step explanation:

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2x^5-x^2+1=0<br> can you help me ?<br> slove it in details<br> thanks

netineya [11]

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Hi my lil bunny!

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$\boxed{x = 7}$

Move 1 to the left side of the equation by subtracting it from both sides.

$\sqrt{2x -5 - 2 - 1 = 0 }$

Subtract 1 from -2.

$\sqrt{2x -5 - 3 = 0 }$

Add 3 to both sides of the equation.

$\sqrt{2x - 5 = 3}$

To remove the radical on the left side of the equation, square both sides of the equation.

$\sqrt{2x - 5^3 = 3^2}$

Simplify each side of the equation.

Multiply the exponents in $(( 2x - 5) ^\frac{1}{2})^2$ .

Apply the power rule and multiply exponents, $(a^m)^n = a^mn$

$(2x -5)^\frac{1}{2}.2 = 3^2$

Cancel the common factor of 2.

$(2x - 5)^1 = 3^2$

Simplify.

$2x - 5 = 3^2$

Raise 3 to the power of 2.

$2x - 5 = 9$

Solve for x

Move all terms not containing x to the right side of the equation.

Add 5 to both sides of the equation.

$2x = 9 + 5$

Add 9 and 5.

$2x = 14$

Divide each term by 2 and simplify.

Divide each term in 2x = 14 by 2.

$\frac{2x}{2} = \frac{14}{2}$

Cancel the common factor of 2.

$x = \frac{14}{2}$

Divide 14 by 2.

$x = 7$

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Hope this helped you.

Could you maybe give brainliest..?

❀*May*❀

6 0

3 years ago

Read 2 more answers

Help me please asap thank you

Korolek [52]

Answer:

3/5

Step-by-step explanation:

The line is going up so it is positive and it is over 3 and up 5 so it would be 3/5.

Hope this helps! Sry if i get it wrong!

3 0

2 years ago

Write a function rule using function notation that will transform a geometric figure by rotating it 270 degrees clockwise. a f(x

zheka24 [161]

Answer:

Step-by-step explanation:

Under a clockwise rotation about the origin of 270°

a point (x, y ) → (- y, x ) , thus

f(x, y ) = (- y, x )

8 0

3 years ago

Find the equation of the problem

pogonyaev

$6\sin^2\theta-\sin\theta=1\ \ \ -1\\\\6\sin^2\theta-\sin\theta-1=0\\\\6\sin\theta-3\sin\theta+2\sin\theta-1=0\\\\3\sin\theta(2\sin\theta-1)+1(2\sin\theta-1)=0\\\\(2\sin\theta-1)(3\sin\theta+1)=0\iff2\sin\theta-1=0\ \vee\ 3\sin\theta+1=0\\\\2\sin\theta=1\ \ |:2\ \ \vee\ \ 3\sin\theta=-1\ \ |:3\\\\\sin\theta=\dfrac{1}{2}\ \vee\ \sin\theta=-\dfrac{1}{3}\\\\$
$\theta=30^o\ \vee\ \theta=150^o\ \vee\ \theta=360^o-\sin^{-1}\left(\dfrac{1}{3}\right)\ \vee\ \theta=180^o+\sin^{-1}\left(\dfrac{1}{3}\right)$

5 0

3 years ago

Help me please! What is the solution to this equation?

marusya05 [52]

Answer:

A. 21...............

8 0

2 years ago

Read 2 more answers