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4 years ago

Which of the following points lies on the circle whose center is at the origin and whose radius is 5?

2 answers:

8 0

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Equation of the circumference whose center is at $\mathsf{C(0,\,0)}$ and whose radius is $\mathsf{r=5:}$

$\mathsf{(x-x_C)^2+(y-y_C)^2=r^2}\\\\
\mathsf{(x-0)^2+(y-0)^2=5^2}\\\\
\mathsf{x^2+y^2=25}$

We are looking for the point that satisfies the equation above.

 • Testing $\mathsf{(-3,\,4):}$

$\mathsf{(-3)^2+4^2}\\\\
\mathsf{9+16}\\\\
\mathsf{25\qquad\quad\checkmark}$

The point $\mathsf{(-3,\,4)}$ lies in the circumference.

 • Testing $\mathsf{(1,\,-2):}$

$\mathsf{1^2+(-2)^2}\\\\ \mathsf{1+4}\\\\ \mathsf{5\ne 25\qquad\quad\diagup\hspace{-9}\diagdown}$

The point $\mathsf{(1,\,-2)}$ doesn't lie in the circumference.

 • Testing $\mathsf{(\sqrt{5},\,\sqrt{5}):}$

$\mathsf{(\sqrt{5})^2+(\sqrt{5})^2}\\\\ \mathsf{5+5}\\\\ \mathsf{10\ne 25\qquad\quad\diagup\hspace{-9}\diagdown}$

So, the point $\mathsf{(\sqrt{5},\,\sqrt{5})}$ doesn't lie in the circumference either.

Answer: (–3, 4).

I hope this helps. =)

vladimir2022 [97]4 years ago

4 0

Answer:

(-3,4)

Step-by-step explanation:

sos

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Answer:

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4 years ago

After heating up in a teapot, a cup of hot water is poured at a temperature of 210°F. The cup sits to cool in a room at a temper

Fiesta28 [93]

Newton's Law of Cooling:

$T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}$

$T(t) =$ Temperature given at a time

$t =$ Time

$T_{s} =$ Surrounding temperature

$T_{o}=$ Initial temperature

$e =$ Constant (Euler's number) ≈ 2.72

$k =$ Constant

Using this information, find the value of $k$ , to the nearest thousandth, then use the resulting equation to determine the temperature of the water cup after 4 minutes.

First, plug in the given values in the equation and solve for $k$ :

$T(t) =$ 197°, $t =$ 1.5 minutes, $T_{s} =$ 70° and $T_{o}=$ 210°

$T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}\\197=70+(210-70)e^{-1.5k} \\197 -70 = (140)e^{-1.5k} \\127 =(140)e^{-1.5k}\\\frac{127}{140}=e^{-1.5k} \\ln(\frac{127}{140})=-1.5k\\-0.097=-1.5k\\0.0649 = k$

$k$ ≈ $0.065$

Let the temperature of the water cup after $t = 4$ minutes be $T(t) = x$

Now, let's plug the new time and $k$ constant in the equation and solve for $x$ :

$T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}\\\\\x=70+(210-70})e^{-0.065*4}\\\\x=70+(140})e^{-0.26}, -0.26=-\frac{26}{100}=-\frac{13}{50} \\$

$x=70+(140})e^{-\frac{13}{50}}\\\\$

$x=70+(140})e^{\frac{1}{\frac{13}{50}}\\\\\\$

$x=70+e^{\frac{140}{\frac{13}{50}}\\\\\\$

$x=70+{\frac{140}{\sqrt[50]{e^{13}}}\\$

$x = 70 +\frac{140}{1.3} \\x=70+107.947\\$

$x=177.95$ ≈ $178$

Temperature of water after 4 minutes is 178°

sorry if there's any misspelling or wrong step but I hope my answer is correct ':3

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3 years ago

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balu736 [363]

Answer:

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3 years ago

Batman (mass=100 kg) jumps straight down from a bridge in to a boat (mass = 580 kg) in which a criminal is fleeing. The velocity

Crank

Answer:

$v_b=11.94\ m/s$

Step-by-step explanation:

Linear Momentum

The momentum of a system of masses m1,m2,m3... with speeds v1,v2,v3,... is given by

$M=m_1v_1+m_2v_2+m_3v_3+...$

When some interactions take place into the system of masses with no external forces interferring, the total momentum is not changed, which means that if the new speeds are v1', v2', v3'... then:

$m_1v_1+m_2v_2+m_3v_3+...=m_1v_1'+m_2v_2'+m_3v_3'+...$

The problem relates the story of Batman (m1=100kg) who is initially assumed at rest and lands on a boat with mass m2=580 kg initially moving at 14 m/s to the positive reference. When the collision takes place, both masses join and a common speed $v_b$ is achieved by the common mass. Applying the conservation of momentum, we have