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Sonja [21]
3 years ago
6

Some help! I just need help with this quick question, God Bless you

Mathematics
1 answer:
Viefleur [7K]3 years ago
8 0
I’m sorry I don’t know. But God Bless you to!
You might be interested in
find the distance between P1 (3,-195degrees) and P2 (-4, -94 degrees) on the polar plane. Round your answer to the nearest thous
Ronch [10]
This is distance formula is sqr. root of (x1-x2)^2 + (y1-y2)^2 
(3-(-4))^2+ (-94-(-195))
 7^2                101^2
  49+                10,201
square root of 10250
101.242 or the closest answer to this

8 0
3 years ago
Two angles have measures of 53° and 127°. What type of angle pair do they form?
vodomira [7]
If it’s a triangle it’s An obtuse angle because so far 53 + 127 equals 180 and obviously if it’s a triangleyou need a third angle so it’d be over 180 meaning obtuse. If those are the only sides and it’s not a triangle then it’s a straight angle because a straight angle is 189 degress
6 0
3 years ago
3(m - n + 2) + 6 - m​
julia-pushkina [17]

Step-by-step explanation:

8 0
3 years ago
The lines below are parallel. If the slope of the solid line is –3, what is the slope of the dashed line?
stepladder [879]

Answer:

-3

Step-by-step explanation:

parallel lines always have the same slope, the y-intercepts will differ

8 0
3 years ago
Find the distance traveled by a particle with position (x, y) as t varies in the given time interval. x = 5 sin2 t, y = 5 cos2 t
rodikova [14]
\begin{cases}x(t)=5\sin2t\\y(t)=5\cos2t\end{cases}\implies\begin{cases}\frac{\mathrm dx}{\mathrm dt}=10\cos2t\\\frac{\mathrm dy}{\mathrm dt}=-10\sin2t\end{cases}

The distance traveled by the particle is given by the definite integral

\displaystyle\int_C\mathrm dS=\int_0^{3\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt

where C is the path of the particle. The distance is then

\displaystyle\int_0^{3\pi}\sqrt{100\cos^22t+100\sin^22t}\,\mathrm dt=10\int_0^{3\pi}\mathrm dt=30\pi
6 0
3 years ago
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