Ok assuming that when he buys the fabric, he cuts pieces off to make it fit, but doesn't sow them in other places to make up the lost area
so
first find the area of tablecloth he needs to buy
area=pir^2
diameter=32
d=2r
r=d/2
r=32/2=16
adds 5 inches extra
16+5=21
r=21
area of cloth=pi21^2
are of cloth=441pi
aprox pi=3.14
area=441 tmes 3.14
area=1384.74
buys cloth with width of 50 and legnth of x
so we need 2 things
1. area of cloth bought>area of tablecloth
2. legnth &width must both be greater or equal to the diameter of the tablecloth (so we can make a circle fit)
new diameter=32+5+5=42 (2 sides)
so
l and w must both be greater than or equal to 42
50l>1384.74
divide both sides by 50 to find legnth
l>27.6948
but the legnth must be also greater than or equal to 42
so make l=42 for min legnth
area of cloth bought=lw=42 by 50=42 times 50=2100
so we do
area wasted=area bought-area needed
area needed=area of tablecloth=1384.74
area bought=2100
area wasted=2100-1384.74
area wasted=715.26 in^2
answer is wasted area=715.26 square inches
63% Please Give me Brainlyest
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: x 3
Step-by-step explanation:
x3 because you want the remove the fraction and make it a whole number. So x=45 would be the answer but x 3 is the operation.
It will says divided by or took away for subtracting and put in for adding
