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3 years ago

Explain how to write the equations of vertical lines, and why they are written this way

Mathematics

1 answer:

Rom4ik [11]3 years ago

3 0

1. Recall the vertical line equation.

2. Plug in the x that we know.

3. Write down the final equation

Pls mark me brainiest :)

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Find the circumference of a circle whose area is 24 pi ft ^2

lianna [129]

$\bf \textit{area of a circle}\\\\
A=\pi r^2~~
\begin{cases}
r=radius\\
-----\\
A=24\pi 
\end{cases}\implies 24\pi =\pi r^2\implies \cfrac{24\pi }{\pi }=r^2
\\\\\\
24=r^2\implies \boxed{\sqrt{24}=r}\\\\
-------------------------------\\\\
\textit{circumference of a circle}\\\\
C=2\pi r\qquad \qquad \implies C=2\pi\left( \boxed{\sqrt{24}} \right)~~
\begin{cases}
24=2\cdot 2\cdot 2\cdot 3\\
\qquad 2^2\cdot 6
\end{cases}
\\\\\\
C=2\pi \sqrt{2^2\cdot 6}\implies C=2\pi (2\sqrt{6})\implies C=4\sqrt{6}~\pi$

3 0

4 years ago

Point AAA is at {(6, -6)}(6,−6)left parenthesis, 6, comma, minus, 6, right parenthesis and point CCC is at {(-6,-2)}(−6,−2)left

garik1379 [7]

Answer:

B(-3, -3)

Step-by-step explanation:

If a point O(x, y) divides line segment XY in the ratio of n:m and the endpoints of the segment are $X(x_1,y_1)\ and\ Y(x_2,y_2)$ , the coordinates of O is:

$x=\frac{n}{n+m}(x_2-x_1)+x_1 \\\\y=\frac{n}{n+m}(y_2-y_1)+y_1$

Given that A(6, -6) and C(-6, 2). Pont B is on AC such that:

AB = (3/4)AC

AB/AC = 3/4

Therefore point B divides the line AC in the ratio of 3:1. Let point B be at (x, y), therefore:

$x=\frac{3}{3+1}(-6-6)+6=\frac{3}{4}(-12)+6=-9+6=-3\\ \\y=x=\frac{3}{3+1}(-2-(-6))-6=\frac{3}{4}(4)-6=3-6=-3$

Therefore the location of B is at (-3, -3)

8 0

3 years ago

Figure ABCD is plotted on a coordinate plane. The figure transforms to create figure A'B'CD'. Which transformation took place?

bulgar [2K]

Answer: A(3,4) A'(-3,4)

B(5,4) B'(-5,4)

C(4,2) C'(-4,2)

D(2,2) D'(-2,2)

Step-by-step explanation:

5 0

4 years ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

7 0

3 years ago

Based on the provided information about the characteristic roots and the right hand side function g(t), determine the appropriat

Rufina [12.5K]

Answer:

Yp = t[Asin(2t) + Acos(2t)]

Yp = t²[At² + Bt + C]

Step-by-step explanation:

The term "multiplicity" means when a given equation has a root at a given point is the multiplicity of that root.

(a) r1=-2i; r2=2i g(t)=2sin(2t) + 3cos(2t)

As you can notice the multiplicity of this equation is 1 since the roots r1 = 2i and r2 = 2i appear for only once.

The form of a particular solution will be

Yp = t[Asin(2t) + Acos(2t)]

where t is for multiplicity 1

(b) r1=r2=0; r3=1 g(t)= t² +2t + 3

As you can notice the multiplicity of this equation is 2 since the roots r1 = r2 = 0 appears 2 times.

The form of a particular solution will be

Yp = t²[At² + Bt + C]

where t² is for multiplicity 2

6 0

3 years ago