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3 years ago

2x + 3 > 13 Whats the answer plz help

Mathematics

1 answer:

Naily [24]3 years ago

8 0

Answer:

In this case, you can simply treat the > sign like an = sign. The only time you would need to worry about is when multiplying or dividing by negative numbers.

2x + 3 > 13

2x + 3 - 3 > 13 - 3

2x > 10

2x / 2 > 10 / 2

x > 5

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

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

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Solve the trigonometric equation:

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

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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

Help me pleaseeeeeeeeee

ahrayia [7]

for 2 points A(xa;ya) and B(xb;yb)

the slope is (yb - ya) / (xb - xa)

here

the slope = 4 = (17-r) / (4-(-1))

(17-r) / 5 = 4

17 - r = 20

r = 17 - 20

r = - 3

point (-1 ; -3)

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2 years ago