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Tcecarenko [31]
3 years ago
15

IF CORRECT I WILL GIVE BRAINLY I NEED HELP ASAP

Mathematics
2 answers:
Dafna11 [192]3 years ago
8 0
1 question what is x equal to?
zvonat [6]3 years ago
7 0

Answer:

D

Step-by-step explanation:

I got it right on the assignment

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givi [52]

Answer:

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2 years ago
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Simplify the expression.Can you help me with this one. I'm stuck at 12^x4 - 15x^2
Blababa [14]

Simplify the given expression as shown below

\begin{gathered} 3x(4x^4-5x)=3x*4x^4+3x(-5x)=12x^5+(-15x^2) \\ =12x^5-15x^2 \end{gathered}<h2>Therefore, the answer is 12x^5-15x^2</h2>

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1 year ago
If lim x-&gt; infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
MAXImum [283]

We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

So

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

and this happens if <em>a</em> = 1 and <em>b</em> = -1.

To confirm, we have

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0

as required.

3 0
3 years ago
What is the domain of:<br><br> x ≥ -2<br><br> x ≥ 0<br><br> x ≥ 2<br><br> all real numbers
777dan777 [17]
Your answer should be A! 

I hope this helped!

If you have any further questions, please feel free to private message me!
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Are there options? Otherwise the definition is: <span>a figure formed by the intersection of a plane and a right circular cone. Depending on the angle of the plane with respect to the cone, a conic section may be a circle, an ellipse, a parabola, or a hyperbola.
</span>
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3 years ago
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