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Ainat [17]
4 years ago
13

2

Mathematics
1 answer:
natali 33 [55]4 years ago
6 0
2x i think thats the answer i really dont know
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22,8,7,2,-11 these integers in order from greatest to least
Alchen [17]
They are already in order from greatest to least
8 0
3 years ago
Please help me figure this out explain it to me
jeka94

Answer:

40yd

Step-by-step explanation:

area of total minus unshaded portion

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3 years ago
Write the equation of the line passing through the given pair of points. (-9, 7) and (-9, -1)
aksik [14]
The first thing I should do with these two points is find the slope of the line.

m = (y2 - y1)/(x2 - x1)

m = (-1 - 7)/(-9 - -9)
m = -8/0

The slope is undefined. We cannot divide by 0. This makes sense because the points (-9, 7) and (-9, -1) would be right on top of each other.  This is a vertical line that has an x-intercept at -9. 

Your equation would be x = -9.
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3 years ago
15=3/5(6x + 20) solve for x
NISA [10]
This is the answer to the problem. Hope it helps
8 0
3 years ago
The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the veloc
andreev551 [17]

Answer:

\theta= \frac{\pi}{2} +\pi \cdot i, for all i = \mathbb{Z} \cup\{0\}

Step-by-step explanation:

The velocity vector is found by deriving the position vector depending on the time:

\dot r(t)= v (t) = 3 \cdot i +\sqrt{2} \cdot j + 2\cdot t \cdot k

In turn, acceleration vector is found by deriving the velocity vector depending on time:

\ddot r(t) = \dot v(t) = a(t) = 2 \cdot k

Velocity and acceleration vectors at t = 0 are:

v(0) = 3\cdot i + \sqrt{2} \cdot j\\a(0) = 2 \cdot k\\

Norms of both vectors are, respectively:

||v(0)||\approx 3.317\\||a(0)|| \approx 2

The angle between both vectors is determined by using the following characteristic of a Dot Product:

\theta = \cos^{-1}(\frac{v(0) \bullet a(0)}{||v(0)||\cdot ||a(0)||})

Given that cosine has a periodicity of \pi. There is a family of solutions with the form:

\theta= \frac{\pi}{2} +\pi \cdot i, for all i = \mathbb{Z} \cup\{0\}

7 0
3 years ago
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