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3 years ago

On a number line what comes first the Y or the X

Mathematics

1 answer:

zavuch27 [327]3 years ago

7 0

Answer:

Rise/Run = y/x

Step-by-step explanation:

On a number line the X-axis is on the bottom, and the Y-axis is on the Top.

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A(4,-7) B(-2,1) find AB

Natali5045456 [20]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{4}~,~\stackrel{y_1}{-7})\qquadB(\stackrel{x_2}{-2}~,~\stackrel{y_2}{1})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[-2-4]^2+[1-(-7)]^2}\implies AB=\sqrt{(-2-4)^2+(1+7)^2}\\\\\\AB=\sqrt{(-6)^2+8^2}\implies AB=\sqrt{36+64}\implies AB=\sqrt{100}\implies AB=10$

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3 years ago

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Studentka2010 [4]

Answer:

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Step-by-step explanation:

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3 years ago

What ratio can you use to determine the probability of a compound event?

maria [59]

Answer:

Step-by-step explanation:

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4 0

3 years ago

Given an acceleration vector, initial velocity u0,v0,w0 , and initial position x0,y0,z0 , find the velocity and position vectors

dsp73

Answer:

Thus we find that velocity vector at time t is

(5t+15, 5t^2/2, 4t^2)

Step-by-step explanation:

given that acceleration vector is a funciton of time and at time t

$a(t) = (5,5t, 8t)$

v(t) can be obtained by integrating a(t)

v(t) = $(5t, 5t^2/2, 4t^2)+(u_0,v_0,w_0)\\=(5t+15, 5t^2/2, 4t^2)$

Thus we use the fact that acceleration is derivative of velocity and velocity is antiderivative of acceleration.

The arbitary constant normally used for integration C is here C vector = initial velocity (u0,v0,w0)

Position vector can be obtained by integrating v(t)

Thus we find that velocity vector at time t is

(5t+15, 5t^2/2, 4t^2)

5 0

3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago