Answer:
-2?
Step-by-step explanation:
use y2-y1 over x2-x1 to find the answer. So you would do -2--5/3 over 4--2. The two negitives turn into a positive. SO it is now -2+5/3 over 4+2. -2+5/3= -1/3 over 4+2=6 it is now -1/3/6 which equals -2. So I think the answer is -2.<u> I do not know if you should trust me on this but I tried. </u>
Answer: x*3 -6=12
Explanation: I assume you wanted that put into an equation
So legnt=l
width=w
l=-7+3w
l=3w-7
area=66
area=legnth times width or
area=lw
subisutte
area=lw
l=3w-7
66=(3w-7)(w)
66=3w^2-7w
subtrac 66 from both sides
0=3w^2-7w-66
factor
HOW TO
multiply 3 and -66=-198
find what 2 numbers add up to -7 and multiply to get -198
the numbers area 11 and -18
so
3w^2-18w+11w-66=0
factor
(3w^2-18w)+(11w-66)=0
(3w)(w-6)+(11)(w-6)=0
factor using distributive (ab+ac=a(b+c))
(3w+11)(w-6)=0
set them to zero
3w+11=0
subtract 11
3w=-11
legnths cannot be negative so we discard this
w-6=0
add 6
w=6
width=6
legnth=3w-7
length=3(6)-7
legnt=18-7
legnth=11
width=6
legnth=11
Answer:
BD = 12.1 (nearest tenth)
Step-by-step explanation:
∆ABC is an isosceles triangle, since it has two equal sides, AB and BC. Also, this means that <BAD and <BCD = 60° each.
BD divides ∆ABC into two equal parts.
Apply trigonometric ratio to find BD.
Reference angle = <BAD = 60°
Adjacent = AD = 7
Opposite = BD
Thus, we would have:
tan 60 = opp/adj
Tan 60 = BD/7
7*Tan 60 = BD
12.1 = BD
BD = 12.1 (nearest tenth)
The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span> =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).
