The position function of a particle is given by:
The velocity function is the derivative of the position:
The particle will be at rest when the velocity is 0, thus we solve the equation:
The coefficients of this equation are: a = 2, b = -9, c = -18
Solve by using the formula:
![t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Substituting:
![\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81-4%282%29%28-18%29%7D%7D%7B2%282%29%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81%2B144%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B225%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm15%7D%7B4%7D%20%5Cend%7Bgathered%7D)
We have two possible answers:
We only accept the positive answer because the time cannot be negative.
Now calculate the position for t = 6:
Answer:
50 increased by 30% is 65
Step-by-step explanation:
'Percent (%)' means 'out of one hundred':
p% = p 'out of one hundred',
p% is read p 'percent',
p% = p/100 = p ÷ 100.
30% = 30/100 = 30 ÷ 100 = 0.3.
100% = 100/100 = 100 ÷ 100 = 1.
Percentage increase = 30% × 50
New value = 50 + Percentage increase
Calculate
New value =
50 + Percentage increase =
50 + (30% × 50) =
50 + 30% × 50 =
(1 + 30%) × 50 =
(100% + 30%) × 50 =
130% × 50 =
130 ÷ 100 × 50 =
130 × 50 ÷ 100 =
6,500 ÷ 100 =
65
Or just find 30% of 50 (15) and add that
4(3x−1)=3+8x−11
Simplify
4(3x−1)=3+8x−11
(4)(3x)+(4)(−1)=3+8x+−11(Distribute)
12x+−4=3+8x+−11
12x−4=(8x)+(3+−11)(Combine Like Terms)
12x−4=8x+−8
12x−4=8x−8
Subtract 8x from both sides.
12x−4−8x=8x−8−8x
4x−4=−8
Add 4 to both sides.
4x−4+4=−8+4
4x=−4
Divide both sides by 4.
4x/4=−4/4
x=−1
-b +or- root (b^2 -4ac) / 2a
3+or- root 9+20/ 2
3+ or - root 29 /2
5.38516480713 + 3 = 8.385 / 2
x = 4.1925
or x = -1.1925
a=1
b=-3
c=-5
