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3 years ago

What is the limit as x is approaching 0 of the absolute value of x

1 answer:

6 0

Recall that, an absolute value expression is a piece-wise in effect, thus we'd check the single-sided limits,

$\bf \lim\limits_{x\to 0}~|x|\implies 
\begin{cases}
\lim\limits_{x\to 0^-}~-x\\\\
\lim\limits_{x\to 0^+}~+x
\end{cases}\implies 
\begin{cases}
0\\\\
0
\end{cases}\implies \lim\limits_{x\to 0}~|x|=0$

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Select the equivalent expressions for z divided by 144

Ivahew [28]

Answer:

A: 144/z

Step-by-step explanation:

Hopefully this helps!

4 0

2 years ago

C. A rational number is a number that can be expressed as a fraction with both the

Dmitriy789 [7]

Answer:

Step-by-step explanation:

Rational expressions are bascially an expression where one polynomial is divided by another. And a polynomial is a series of terms like ax^b added together, where the exponents, what I labeled as b, need to be positive integers.

6 0

2 years ago

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

Mila [183]

Answer:

$m\angle H = 44.4\°$ .

Step-by-step explanation:

Given:

In Right Angle Triangle GIH

∠ I = 90°

GI = 7 ....Side opposite to angle H

GH = 10 .... Hypotenuse

To Find:

m∠H = ?

Solution:

In Right Angle Triangle ABC ,Sine Identity,

$sin \ H = \frac{Oppsite\ side\ to\ \angle H}{Hypotenuse}$

Substituting the values we get;

$sin\ H = \frac{7}{10} = 0.7$

Now taking $sin^{-1}$ we get;

$\angle H = sin^{-1}\ 0.7 = 44.427$

rounding to nearest tenth we get.

$m\angle H = 44.4\°$ .

Hence $m\angle H = 44.4\°$ .

6 0

2 years ago

A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in and then s

Bas_tet [7]

Answer:

Step-by-step explanation:

$m=\frac{3}{32}\\ \gamma =0\text{ damping constant.}\\ k=\frac{3}{\frac{1}{4}}=12\\ F(t)=0\\ \text{so equation of motion is }\\ mu''+\gamma u'+ku=F(t)\\ \frac{3u''}{32}+12u=0\\ u''+128u=0\\ \text{and initial conditions are}\\ u(0)=-\frac{1}{12}\\ u'(0)=2\\ u''+128u=0\\ \text{characterstic equation is}\\ r^2+128=0\\ \text{roots are }\\ r=\pm 8\sqrt{2}i\\ \text{therefore general solution is}\\ u(t)=A\cos8\sqrt{2}t+Bsin8\sqrt{2}t\\ \text{using initial conditions we get}\\ A=-\frac{1}{12}\\ B=\frac{\sqrt{2}}{8}\\ \text{therefore solution is }\\ u(t)=-\frac{1}{12}\cos8\sqrt{2}t+\frac{\sqrt{2}}{8}\sin8\sqrt{2}t\\
{hence}\\R=\sqrt{\frac{11}{288}}\\\\\sigma=\pi - \tan^{-1}\frac{1}{\sqrt{2}}\\\\\omega_0=8\sqrt{2}\\T=\frac{4}{4\sqrt{2}}$