Answer:
80in²
Step-by-step explanation:
Hey There!
So the first step of this problem is to find the width of the scale drawing
To do so we need to find the scale factor
To find the scale factor we divide the length of the room by the length of the scale factor
32/8=4
so the scale factor is 4
now to find the width
40/4=10
so the width of the scale drawing is 10 in
now we can find the area given the length and the width
10x8=80in²
so we can conclude that the area of the scale drawing is 80in²
Answer:
\frac{-2x^2+2}{x^4+2x^2+1}
2x^2+2
-------------
x^4+2x^2+1
Answer:
Step-by-step explanation:
When I see "at what rate", I know this question must come from
pre-Calculus, so I won't feel bad using a little Calculus to solve it.
-- The runner, first-base, and second-base form a right triangle.
The right angle is at first-base.
-- One leg of the triangle is the line from first- to second-base.
It's 90-ft long, and it doesn't change.
-- The other leg of the triangle is the line from the runner to first-base.
Its length is 90-24T. ('T' is the seconds since the runner left home plate.)
-- The hypotenuse of the right triangle is
square root of [ 90² + (90-24T)² ] =
square root of [ 8100 + 8100 - 4320T + 576 T² ] =
square root of [ 576 T² - 4320 T + 16,200 ]
We want to know how fast this distance is changing
when the runner is half-way to first base.
Before we figure out when that will be, we know that since
the question is asking about how fast this quantity is changing,
sooner or later we're going to need its derivative. Let's bite the
bullet and do that now, so we won't have to worry about it.
Derivative of [ 576 T² - 4320 T + 16,200 ] ^ 1/2 =
(1/2) [ 576 T² - 4320 T + 16,200 ] ^ -1/2 times (576T - 4320) .
There it is. Ugly but manageable.
How fast is this quantity changing when the runner is halfway to first-base ?
Well, we need to know when that is ... how many seconds after he leaves
the plate.
Total time it takes him to reach first-base = (90 ft)/(24 ft/sec) = 3.75 sec .
He's halfway there when T = (3.75 / 2) = 1.875 seconds. (Seems fast.)
Now all we have to do is plug in 1.875 wherever we see 'T' in the big derivative,
and we'll know the rate at which that hypotenuse is changing at that time.
Here goes. Take a deep breath:
(1/2) [ 576 T² - 4320 T + 16,200 ] ^ -1/2 times (576T - 4320) =
[ 576 T² - 4320 T + 16,200 ] ^ -1/2 times (1152T - 8640) =
[576(1.875)² - 4320(1.875) + 16,200]^-1/2 times [1152(1.875)-8640] =
[ 2,025 - 8,100 + 16,200 ] ^ -1/2 times [ 2,160 - 8640 ] =
- 6480 / √10,125 = - 64.4 ft/sec.
I have a strong hunch that this answer is absurd, but I'm not going to waste
any more time on it, (especially not for 5 points, if you'll forgive me).
I've outlined a method of analysis and an approach to the solution, and
I believe both of them are reasonable. I'm sure you can take it from there,
and I hope you have better luck with your arithmetic than I've had with mine.
Answer: If I’m not mistaken, the largest area is 75 m2 and the smallest is 2 m2
Step-by-step explanation:
For both : since they are asking for the area, you should calculate the perimeter.
So for the largest area, since you know that you have 40 fences, and each are 1 meter long, and they’re also asking for the largest rectangular area, this means that only the opposite sides will have the same length, so in order to divide the rectangle with the 40 one-meter fences, the biggest sides have to be 15 meters long (so for both : 15x2=30 meters). Then you deduct from 40, 30 : 40-30 = 10 which then you can devide by 2 and find 5 + 5 and that means that the 2 smallest sides will be 5 meters long each. Then finally to calculate the perimeter you should do the biggest side multiplied by the smallest side : 15 x 5 and you find 75 m2 (squaremeters), if I’m not mistaken.
And for the smallest area you take the smallest possibilities knowing that only the opposite sides should have the same length and not all of them, so you do 1 x 2 and you find 2 m2.
