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zheka24 [161]
3 years ago
7

What is the value of x in the equation? 6(x +1) -5x = 8 + 2(x - 1)

Mathematics
2 answers:
Artyom0805 [142]3 years ago
7 0

Answer:

x = 0

Step-by-step explanation:

sineoko [7]3 years ago
4 0

Answer: x =0

Step-by-step explanation:

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B/16 = B-6/4<br><br> Find B:<br><br> Linear Equalities
mamaluj [8]

Answer:

B=8

Step-by-step explanation:

B/4=B-6/4

4B=16B-96

96=12B

B=8

4 0
3 years ago
What are the vertical asymptotes of the function f(x) =5x+5/x2 + x-2
katen-ka-za [31]

let's recall that the vertical asymptotes for a rational expression occur when the denominator is at 0, so let's zero out this one and check.

\bf \cfrac{5x+5}{x^2+x-2}\qquad \stackrel{\textit{zeroing out the denominator}~\hfill }{x^2+x-2=0\implies (x+2)(x-1)=0}\implies \stackrel{\textit{vertical asymptotes}}{ \begin{cases} x=-2\\ x=1 \end{cases}}

5 0
3 years ago
What is the answer cos75 degrees=10/x
vova2212 [387]
To find the answer we simply work out the equation.
cos (75) = 10/x

cos (75) * x = 10. Here, I simply multiplied both sides by x to move x to the left hand side of the equation.

x = 10 / cos (75) Here, I divided cos (75) on both sides to move cos (75) to the right hand side of the equation.

The cosine of 75 is 0.92175127, so, 10 / 0.92175127 = 10.8489137
4 0
3 years ago
Please please help me!!!
Dafna1 [17]
Answer is 40. A triangle equals 180 so all you do is subtract 65 and 75 from 180 to get your answer.
5 0
3 years ago
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
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