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3 years ago

6

How do I solve?

TexFormula1" title="2 = \frac{x(x-1)(x-3)^{2} (x+1)}{x}" alt="2 = \frac{x(x-1)(x-3)^{2} (x+1)}{x}" align="absmiddle" class="latex-formula">

2 answers:

Rus_ich [418]3 years ago

8 0

Answer:

x = 5

Step-by-step explanation:

x(x - 1)(x - 3)^2(x + 1)

x - x + x - 9 + x + 1

x - 9 + x + 1

2x - 8 = 2

<u> +8 +8</u>

2x = 10

<em><u>x = 5</u></em>

Alexxandr [17]3 years ago

6 0

Answer:

x=5

Step-by-step explanation:

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Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

——————————

Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

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3 years ago

5 (u + 1) -<br>7 = 3<br>3 (u - 1) + 2u

frez [133]

Correct Question:

5 (u + 1) - 7 = 3 (u - 1) + 2u.

Solve for u

Answer:

See explanation below

Step-by-step explanation:

In this given question, we are required to find u.

Given the equation:

5 (u + 1) - 7 = 3 (u - 1) + 2u

Required:

Solve for u

To find u, first simplify both sides individually.

Simply 5 (u + 1) - 7:

Expand the parenthesis:

5u + 5 - 7

Collect like terms:

5u - 2

<em>Simplify 3 (u - 1) + 2u:</em>

Expand the parenthesis:

3u - 3 + 2u

Collect like terms:

3u + 2u - 3

5u - 3

Bring both simplified equations together:

5u - 2 = 5u - 3

5u - 5u - 2 = -3

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Since -2 ≠ -3, there is no solution.

Therefore, we can say the equation is invalid.

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3 years ago

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Hoochie [10]

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Step-by-step explanation:

I will suppose that this is not a continuos probability, as the individual probabilites add up to 100%.

If you want to obtain the probability that x ≤ 0, then you need to add the probability for the cases x= 0, x = -1, x = -2 .... etc

This is:

x = 0, p = .16

x = -2, p = .33

x = -3, p = .13

x = -5, p = .17

Then, the probability where x takes a negative value or zero {-5, -3, -2, 0} is:

P = 0.16 + 0.33 + 0.13 + 0.17 = 0.79

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3 years ago