Slope of line = tan(120) = -tan(60) = - √3
Distance from origin = 8
Let equation be Ax+By+C=0
then -A/B=-√3, or
B=A/√3.
Equation becomes
Ax+(A/√3)y+C=0
Knowing that line is 8 units from origin, apply distance formula
8=abs((Ax+(A/√3)y+C)/sqrt(A^2+(A/√3)^2))
Substitute coordinates of origin (x,y)=(0,0) =>
8=abs(C/sqrt(A^2+A^2/3))
Let A=1 (or any other arbitrary finite value)
solve for positive solution of C
8=C/√(4/3) => C=8*2/√3 = (16/3)√3
Therefore one solution is
x+(1/√3)+(16/3)√3=0
or equivalently
√3 x + y + 16 = 0
Check:
slope = -1/√3 .....ok
distance from origin
= (√3 * 0 + 0 + 16)/(sqrt(√3)^2+1^2)
=16/2
=8 ok.
Similarly C=-16 will satisfy the given conditions.
Answer The required equations are
√3 x + y = ± 16
in standard form.
You can conveniently convert to point-slope form if you wish.
Answer:
Given,
p (x)=2x-3
g (x)=6x+4
p (x)×g (x)=?
Step-by-step explanation:
p (x)×g (x)=(2x-3)(6x+4)
=2x (6x+4)-3 (6x+4)
=12x^2+8x-18x-12
=12x^2-10x-12
Answer:
could you ask in english
Step-by-step explanation:
Answer:
Total = 686000 meters for 10 weeks 7 days a week.
Step-by-step explanation:
Let's solve this by doing the increase on just 1 training day per week. Then we can find the grand total for 7 days a week by multiplying by 7
a = 8000
n = 10
d = 400
Sum = (a + L)*n/2
- L = a + (n - 1)*d
- L = 8000 + (10 - 1)*400
- L = 8000 + 9*400
- L = 8000 + 3600
- L = 11600 meters
Sum = (8000 + 11600)*10/2
Sum = 19600 * 10/2
Sum = 98000 meters.
This total represents 1 training day a week.
The total amount is 7 times this figure.
Total = 7 * 98000 = 686 000 meters.
