What is the domain of the relation shown?

What is 20-14+7-9+6-2+24=?

Rashid [163]

Answer:

32

Step-by-step explanation:

5 0

2 years ago

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ΔABC is a right triangle in which is a right angle, AB = 1, AC = 2, and BC = .

ollegr [7]

We are asked to solve for the measurement of side BC in the given right triangle ΔABC and other side measurements were also given such as AB=1 and AC = 2. Since this is a right triangle, we can use and apply the Pythagorean theorem c²= a² + b² and the solution is shown below:
c = AC
b = BC
a = AB
AC² = AB² + BC² , substitute values we have:
2² = 1² + BC²
BC² = 4-1
BC = √3
BC = 1.732

The answer for the length of BC is 1.732 units.

8 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:

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