Answer:
The perpendicular line 4x − 5y = 20 has a slope of 4/5 and as such, the slope of the required line = -5/4
(y - 3)/(x + 4) = -5/4 simplify to get the required equation: 5x + 4y + 8 = 0
Answer:
Answer is in the attachment.
Step-by-step explanation:
To graph x>2 consider first x=2. x=2 is a vertical line and if you want to graph x>2 you need to shade to the right of the vertical line.
To graph x+y<2, I will solve for y first.
x+y<2
Subtract x on both sides:
y<-x+2
Consider the equation y=-x+2. This is an equation with y-intercept 2 and slope -1 or -1/1. So the line you have in that picture looks good for y=-x+2. Now going back to consider y<-x+2 means we want to shade below the line because we had y<.
Now where you see both shadings will be intersection of the shadings and will actually by your answer to system of inequalities you have. In my picture it is where you have both blue and pink.
I have a graph in the picture that shows the solution.
Also both of your lines will be solid because your question in the picture shows they both have equal signs along with those inequality signs.
Just in case my one graph was confusing, I put a second attachment with just the solution to the system.
Try to relax. Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before. But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.
Consider this: (2)^a negative power = (1/2)^the same power but positive.
So:
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.
What I just said in that paragraph was: log₂ of(N) = <em>- </em>log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.
Now let's look at the problem:
log₂(x-1) + log(base 1/2) (x-2) = log₂(x)
Subtract log₂(x) from each side:
log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0
Subtract log(base 1/2) (x-2) from each side:
log₂(x-1) - log₂(x) = - log(base 1/2) (x-2) Notice the negative on the right.
The left side is the same as log₂[ (x-1)/x ]
==> The right side is the same as +log₂(x-2)
Now you have: log₂[ (x-1)/x ] = +log₂(x-2)
And that ugly [ log to the base of 1/2 ] is gone.
Take the antilog of each side:
(x-1)/x = x-2
Multiply each side by 'x' : x - 1 = x² - 2x
Subtract (x-1) from each side:
x² - 2x - (x-1) = 0
x² - 3x + 1 = 0
Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .
I think you have to say that <em>x=2.618</em> is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.
There,now. Doesn't that feel better.
300ft cause A=lw and it said its a square so all sides are the sam
so
300 times 300 = 900
The answer is Hx = ½ Wsin θ cos θ
The explanation for this is:
Analyzing the torques on the bar, with the hinge at the axis of rotation, the formula would be: ∑T = LT – (L/2 sin θ) W = 0
So, T = 1/2 W sin θ. Analyzing the force on the bar, we have: ∑fx = Hx – T cos θ = 0Then put T into the equation, we get:∑T = LT – (L/2 sin θ) W = 0