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denis-greek [22]
3 years ago
8

3x-24 x+10 what is the value of x

Mathematics
2 answers:
Arisa [49]3 years ago
8 0
3x-24=x+10
3x-x=24+10
2x=34
x=34:2=17
Vera_Pavlovna [14]3 years ago
5 0
3x-24=x+10
3x-x=24+10
2x=34
x=34:2=17 hope this helps you in the long run

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Step-by-step explanation:


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For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answe
Rus_ich [418]

To solve the inequality \frac{x^2+x+3}{2x^2+x-6}\ge \:0

\mathrm{Factor\:the\:left\:hand\:side\:}\frac{x^2+x+3}{2x^2+x-6}:

The numerator x^2+x+3 is not factorizable.

so factor the denominator 2x^2+x-6:

2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\

Now take \mathrm{Factor\:out\:common\:term\:}\left(2x-3\right)

Then we get factor of the denominator as \left(2x-3\right)\left(x+2\right)

Thus \frac{x^2+x+3}{2x^2+x-6}=\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}

Now \mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}\\

Signs of x^2+x+3>0

\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0

x\frac{3}{2} is the required solution of the given inequality.

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tresset_1 [31]

Answer:

you would use 85 to find angle 4, because they would add up to 180, so you would subract 85 from 180 to get angle 4

Step-by-step explanation:

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