<span>A) 2a + 3b = 12
B) ab = 6 solving for a
B) a = 6 / b then we substitute this into equation A)
</span><span>A) 12 / b + 3b = 12 </span><span>multiplying this by "b"
A) 12 + 3b^2 = 12b
A) 3b^2 -12b +12= 0 dividing by "3"
A) b^2 -4b + 4 = 0
Factoring
(b-2) * (b-2) = 0
b = 2
Since b = 2 then a = 3
</span>NOW, we put these numbers into:
<span>8a^3 +27b^3
</span>
8*3*3*3 + 27*2*2*2
216 + 216
The answer is 512
Answer:
13/12
Step-by-step explanation:
B - 7/4 = -2/3
B = -2/3 + 7/4
B = 13/12
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.
Step-by-step explanation:
The magnitude of x is equal or less than 7.
=> We have |x| <= 7.
