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1 year ago

Find the distance between A (2,6) and N (5, 10). Round To the nearest tenth

Mathematics

1 answer:

Svetllana [295]1 year ago

7 0

Answer:

5.0 units

Step-by-step explanation:

The distance between two points can be expressed in formula of:

$\displaystyle{d = \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}}$

Determine:

$\displaystyle{(x_2,y_2)}$ is (5,10).
$\displaystyle{(x_1,y_1)}$ is (2,6).

Therefore, substitute in the formula to find distance:

$\displaystyle{d = \sqrt{\left(5-2\right)^2+\left(10-6\right)^2}}\\\\\displaystyle{d = \sqrt{3^2+4^2}}\\\\\displaystyle{d = \sqrt{9+16}}\\\\\displaystyle{d = \sqrt{25}}\\\\\displaystyle{d = 5}$

Therefore, the distance between two points is 5.0 units.

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3 years ago

A boat traveling at a constant rate. In 4 seconds, the boat travels 80 feet what is the rate the boat is traveling at in feel pe

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Step-by-step explanation:

Do 80 divided by 4

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3 years ago

Let B be the basis of P3 consisting of the Hermite polynomials in Exercise 21, and let p.t / D 7 ! 12t ! 8t 2 C 12t 3. Find the

Anastasy [175]

To calculate the relative vector of B we have to:

$P_B=\left[\begin{array}{ccc}3\\3\\-2\\3/2\end{array}\right]$

The coordenates of:

$p(t)= 12t^3-8t^2-12t+7$ , with respect to B satisfy:

$C_1(1)+C_2(2t)+C_3(-2+4t^2)+C_4(-12t+8t^3)= 7-12t-8t^2+12t^3$

Equating coefficients of like powers of t produces the system of equation:

$\left \{ {C_1-2C_3=7} \atop {2C_2-12C_4=-12} \right. \\\left \{ {{4C_3=-8} \atop {8C_4=12}} \right.$

After solving this system, we have to:

$C_1=3\\C_2= 3\\C_3= -2\\C_4= \frac{3}{2}$

And the result is:

$P_B=\left[\begin{array}{ccc}3\\3\\-2\\3/2\end{array}\right]$

Learn more: brainly.com/question/16850761

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3 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

3 years ago

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denis23 [38]

Answer: the correct options are b and d.

Step-by-step explanation:

Let us first define what mutually exclusive events are. If two events are mutually exclusive, it means that they cannot happen at the same time. For example, getting a head and a tail at the same time are mutually exclusive.

Considering a single spin of the spinner, the events that are mutually exclusive are

b)landing on a shaded sector and landing on a 3. This is because 3 is unshaded. You can either land on 3 or an unshaded sector at a time

d)landing on an unshaded sector and landing on a number less than 2. This is because the numbers in the unshaded sectors are greater than 2.

3 0

3 years ago