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Savatey [412]
3 years ago
14

Pls help with math queestion !!!!

Mathematics
1 answer:
salantis [7]3 years ago
5 0

Answer:

-4.66667 (the 6. is repeating)

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Find the velocity v(t)v(t) and speed ∥v(t)∥∥v(t)∥ of a particle whose motion is described by x=6t3−9t2,y=3,z=2t2−4t+2
Nesterboy [21]
Let \mathbf p(t)=(x(t),y(t),z(t)) denote the particle's position. Then its velocity is given by

\mathbf v(t)=\dfrac{\mathrm d\mathbf p}{\mathrm dt}=(x'(t),y'(t),z'(t))
\mathbf v(t)=(18t^2-18t,0,4t-4)

The speed is then given by the absolute value/magnitude/norm of the velocity function,

\|\mathbf v(t)\|=\sqrt{(18t^2-18t)^2+(4t-4)^2}
7 0
3 years ago
Please help.<br> geometry unit 7 lesson 6 for connections...
RUDIKE [14]
1] Square; Rectangle; Quadrilateral.

2] All rectangles are Quadrilaterals.

3] Arrange four equal length sides, so the diagonals are equal length also.

4] As per the Isosceles trapezoid.
Base angles are equal to each other and upper both angles are equal to each other too.
So,
20x+9 = 14x+15
20x-14x = 15-9
6x = 6
x = 1°

K = M = 29°

Sum of all angles of quadrilateral equal to 360°.

360-(29+29) = 360-58 = 302°

Sum of upper both equal angles is 302°.
One of them will be 302/2 = 151°

Answer will be 151°.
8 0
3 years ago
Y-1 = 2(x - 2)?<br> Graph
Firdavs [7]

Answer:graph it and you get 5

Step-by-step explanation:

5 0
3 years ago
Plz need help i cant do it​
tatuchka [14]

Answer

adding 2

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Could someone help me please???​
mylen [45]

3 hours worked - $19.50

0.5 hours - x

x= 0.5(19.5)/3

x=3.25

and then do the same for the other cases

6 0
4 years ago
Read 2 more answers
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