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2 years ago

What is the solution of the equation √2x-3-3=6? A) 42 B) 39 C) 3 D) 6

Mathematics

1 answer:

Firlakuza [10]2 years ago

4 0

Answer:

Step-by-step explanation:

√2x-3-3=6

√2x-3=6+3

√2x-3=9

2x-3=9^2

2x=81+3

X=84/2

X=42

So value of X is 42

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IRINA_888 [86]

X= 49 because the square root of 49 IS 7. :D
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2 years ago

Given sin theta= 6/11 and sec theta < 0, find cos theta and tan theta.

bekas [8.4K]

Answer: option a.

Step-by-step explanation:

By definition, we know that:

$cos^2(\theta)=1-sen^2(\theta)\\\\tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

Substitute $sin(\theta)=\frac{6}{11}$ into the first equation, solve for the cosine and simplify. Then, you obtain:

$cos(\theta)=\±\sqrt{1-(\frac{6}{11})^2}\\\\cos(\theta)=\±\sqrt{\frac{85}{121}}\\\\ cos(\theta)=\±\frac{\sqrt{85}}{11}$

As $sec\theta$ then $cos\theta$ :

$cos(\theta)=-\frac{\sqrt{85}}{11}$

Now we can find $tan\theta$ :

$tan\theta=\frac{\frac{6}{11}}{-\frac{\sqrt{85}}{11}}\\\\tan\theta=-\frac{6\sqrt{85}}{85}$

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3 years ago

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Zina [86]

Answer:

Maby b

Step-by-step explanation:

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2 years ago

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Rashid [163]

The answer is quadrilaterals. d

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3 years ago

Can anyone help me out in my homework. I would love to talk about these problems

katen-ka-za [31]

Step-by-step explanation:

(b) If $F_G = m_1g,$ then

$m_1g = G\dfrac{m_1m_2}{r^2}$

Note that $m_1$ cancel out so we get

$g = G\dfrac{m_2}{r^2}$

Solving for $m_2,$ we get

$m_2 = \dfrac{gr^2}{G}$

$F_G = G\dfrac{m_1m_2}{(2r)^2} = \dfrac{1}{4}\left(G\dfrac{m_1m_2}{r^2}\right)$

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2 years ago