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3 years ago

3x - 3(x+1) - 3 Solution for x

Mathematics

1 answer:

sertanlavr [38]3 years ago

7 0

Answer:

-6

Step-by-step explanation:

3x-3(x+1)-3

3x-3x-3-3

-3-3

-6

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-4y+4=10x standard form ik its 10x+4y=4 but you need to simplify more please help i need to get it in an hr

LuckyWell [14K]

Answer:

x= (2/5)-(2y/5) (Im not sure if this is what you need but this is the answer simplified)

Step-by-step explanation:

Move all terms that don't contain x to the right side and solve.

7 0

2 years ago

Can someone please help me with this

Anit [1.1K]

180 = 96 + 2x + x + 12

180 = 96 + 3x + 12

180 = 108 + 3x

-108 -108

72 = 3x

/3 /3

24 = x

7 0

3 years ago

Read 2 more answers

PLEASE MATH HELP WILL GIVE BRAINLIEST!!

solong [7]

I'm sorry!!!!
where is the x?!

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4 years ago

Read 2 more answers

A car rental company’s standard charge includes an initial fee plus an additional fee for each mile driven. the standard charge

kaheart [24]

Answer:

C=0.55M+24.65

Step-by-step explanation:

0.40M+0.15M=0.55M

18.95+5.70=24.65

3 0

3 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

7 0

3 years ago