Express the number as a ratio of integers. 3.469 = 3.469469469
Answers
3.469469469....
then:
Let x=3.469469469.....
then
1000x-x=(3469.469469469...) - (3.469469469....)
999x = 3466
x=3466 / 999
= 3+469/999
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4 connected lines are shown. A line from point S goes slightly down and to the left to point T to form S T. A line from point T goes slightly down and to the right to point U to form T U. A line from point U goes slightly down and to the left to point V to form U T. A line goes slightly down and to the right to point W to form point W.
Complete the two-column proof.
PLEASE HURRYY
Answers
Answer:
Step-by-step explanation:
Given: ∠T ≅ ∠V; ST || UV
Prove: TU || VW
4 connected lines are shown. A line from point S goes slightly down and to the left to point T to form S T. A line from point T goes slightly down and to the right to point U to form T U. A line from point U goes slightly down and to the left to point V to form U T. A line goes slightly down and to the right to point W to form point W.
Complete the two-column proof.
♣ =
✔ alternate interior angles theorem
♦ =
✔ transitive property
♠ =
✔ converse alternate interior angles theorem
just did the assignment
Answer:
Because if ∠T ≅ ∠V, and ST || UV,
the line will go down degrees to form TU || VW.
Answers
Answer:
Yes additive inverse is two complete opposite numbers if added = 0
Answer:
additive
Step-by-step explanation:
Because the point is not past the postive live or below the negative.
If GC = 10x – 6, and NC = 3x + 5.
Find the length of line GC.
x=
GC=
Answers
Answers
Answer:
fafafsghdjdbdbdgsysvdve. wwmsk
Answers
The particle passes through the origin at t = 0 and t = ±√20. The particle is instantaneously motionless at t = 0 and t = ±√10.
(a) To determine the times at which the particle passes through the origin, we need to find when the position function equals zero. So, we set s(t) = 0 and solve for t.
t4 - 20t2 = 0
Factoring out a t2, we get:
t2(t2 - 20) = 0
Setting each factor equal to zero and solving for t gives us the following solutions:
t = 0 (giving us the initial position), and t = ±√20 (approximately t = ±4.47).
(b) To determine when the particle is instantaneously motionless, we need to find when the velocity of the particle is equal to zero. The velocity function of the particle is the derivative of the position function. So, we differentiate s(t) with respect to t to find the velocity function.
v(t) = s'(t) = 4t³ -40t
Setting v(t) = 0, we have:
4t³ -40t = 0
Factoring out a 4t, we get:
4t(t² - 10) = 0
Setting each factor equal to zero and solving for t gives us the following solutions:
t = 0 (giving us the initial velocity), and t = ±√10 (approximately t = ±3.16).
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Final answer:
The particle passes through the origin at t = 0 and t = √20 seconds. The particle is instantaneously motionless at t = 0 and t = ±√10 seconds.
Explanation:
The position of the particle at time t is given by the equation s(t) = t4 - 20t2. To determine the times when the particle passes through the origin, we set s(t) equal to zero and solve for t. This gives us the quadratic equation t4 - 20t2 = 0, which can be factored as t2(t2 - 20) = 0. The solutions to this equation are t = 0 and t = ±√20. Since t cannot be negative in this scenario, the particle passes through the origin at t = 0 and t = √20 seconds.
To determine the times when the particle is instantaneously motionless, we need to find the times when the velocity of the particle is equal to zero. The velocity of the particle can be found by taking the derivative of the position function with respect to time, v(t) = 4t3 - 40t. Setting this equation equal to zero and solving for t gives us the cubic equation 4t3 - 40t = 0. This equation can be factored as 4t(t2 - 10) = 0. The solutions to this equation are t = 0 and t = ±√10. Therefore, the particle is instantaneously motionless at t = 0 and t = ±√10 seconds.
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Answers
Answer:
q = 0.105uC
Step-by-step explanation:
We can determine the force on one ball by assuming two balls are stationary, finding the E field at the lower right vertex and calculate q from that.
Considering the horizontal and vertical components.
First find the directions of the fields at the lower right vertex. From the lower left vertex the field will be at 0° and from the top vertex, the field will be at -60° or 300° because + charge fields point radially outward in all directions. The distances from both charges are the same since this is an equilateral triangle. The fields have the same magnitude:
E=kq/r²
Where r = 20cm
= 20/100
= 0.2m
K = 9.0×10^9
9.0×10^9 × q /0.2²
9.0×10^9/0.04
2.25×10^11 q
These are vector fields of course
Sum the horizontal components
Ecos0 + Ecos300 = E+0.5E
= 1.5E
Sum the vertical components
Esin0 + Esin300 = -E√3/2
Resultant = √3E at -30° or 330°
So the force on q at the lower right corner is q√3×E
The balls have two forces, horizontal = √3×E×q
and vertical = mg, therefore if θ is the angle the string makes with the vertical tanθ = q√3E/mg
mg×tanθ = q√3E.
..1
Then θ will be...
Since the hypotenuse = 80cm
80cm/100
= 0.8m
The distance from the centroid to the lower right vertex is 0.1/cos30 =
0.1/0.866
= 0.1155m
Hence,
0.8×sinθ = 0.1155
Sinθ = 0.1155/0.8
Sin θ = 0.144375
θ = arch sin 0.144375
θ = 8.3°
From equation 1
mg×tanθ = q√3E
g = 9.8m/s^2
m = 3.0g = 0.003kg
0.003×9.8×tan(8.3)
0.00428 = q√3E
0.00428 = q×1.7320×E
Where E=kq/r²
Where r = 0.2m
0.0428 = kq^2/r² × 1.7320
K = 9.0×10^9
0.0428/1.7320 = 9.0×10^9 × q² / 0.2²
0.02471×0.04 = 9.0×10^9 × q²
0.0009884 = 9.0×10^9 × q²
0.0009884/9.0×10^9 = q²
q² = 109822.223
q = √109822.223
q = 0.105uC