Answer:
Ms, Erickson added 6x^3 plus -3x^2 and these terms have different exponents.l
Step-by-step explanation:
This equation uses two properties of logarithms:
So you could take the ln from left and right hand side in the equation, and get:
(2-x)ln 3 = x ln 5
then
2 ln 3 - x ln 3 - x ln 5 = 0 =>
x(ln 3 + ln 5) = 2 ln 3
so x = 2 ln 3 / (ln3 + ln5)
Now using the 1st property you can say 2 ln 3 is ln 3² = ln9
and using the 2nd property you can say ln3 + ln5 = ln15
so x= ln9 / ln15
Answer:
27x +42+ 3x²
Step-by-step explanation:
so im learning this rn in class and what you do is FOIL
So its stands for
First
Outside
Inside
Last
(3x+6)(x+7)
(3x+6)(x+7) -- 3x² (if you have x times x it is x to the second power)
(3x+6)(x+7) -- 21x
(3x+6)(x+7) -- 6x
(3x+6)(x+7) -- 42
Now add 21x and 6x
27x +42+ 3x²
you must have added the 21 and 6 incorrectly.
Answer:
Both are 72 do you need steps?
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.
