Answer:
yes
Step-by-step explanation:
they both multiply by 20
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.
H=-16t² +v(0)t+h(0)
v(0)=192
h=-16t² +192t+h(0)
h(0) should be 0, because the mortar sits on the ground.
h= - 16t²+192t
This function will have maximum because it has minus before x², and parabola is looking down.
h=-(16t²-192t)=-(16t² -2*4t*24+24²)+24²
h=-(4t-24)²+24²
h=-(4t-24)²+576
vertex (24 feet, 576 feet)
24 feet horizontally from the mortar and 576 feet up
Answer:
1530 < 12530
Step-by-step explanation:
i12530 is a bigger number than 1530
Answer:
is it 8x
i'm not sure
Step-by-step explanation:
