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Sergeeva-Olga [200]
3 years ago
5

Can u guys help me with #1 plz guys now

Mathematics
1 answer:
myrzilka [38]3 years ago
3 0
3

the area is 41.6cm which is 1foot  4 3/8 inches.

So 6in + 6in + 6in = 1ft 6in, which will be enough to cover the floor.
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R x 9

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Help me please 10 points I’m only in 8th
AnnyKZ [126]

The answer is

5.09902

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The soccer club has planned a trip to a tournament. The cost of the van is $192. When 2 students who are not members of the club
Anon25 [30]
Let
N--------> <span>members of the club

we know that
($192/(N+2)=3.20
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Can someone help me out with this
katen-ka-za [31]

Answer:

AB=4.41

Step-by-step explanation:

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cos25=\frac{4}{x}

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x*(cos25)=x*(\frac{4}{x})\\\\x*cos25=4

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Insert into a calculator:

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5 0
3 years ago
Using differential calculus, maximize the volume of a box made of cardboard (top is open) as shown in Figure A. 15, subject to t
lana [24]

The maximum volume of the box is 40√(10/27) cu in.

Here we see that volume is to be maximized

The surface area of the box is 40 sq in

Since the top lid is open, the surface area will be

lb + 2lh + 2bh = 40

Now, the length is equal to the breadth.

Let them be x in

Hence,

x² + 2xh + 2xh = 40

or, 4xh = 40 - x²

or, h = 10/x - x/4

Let f(x) = volume of the box

= lbh

Hence,

f(x) = x²(10/x - x/4)

= 10x - x³/4

differentiating with respect to x and equating it to 0 gives us

f'(x) = 10 - 3x²/4 = 0

or, 3x²/4 = 10

or, x² = 40/3

Hence x will be equal to 2√(10/3)

Now to check whether this value of x will give us the max volume, we will find

f"(2√(10/3))

f"(x) = -3x/2

hence,

f"(2√(10/3)) = -3√(10/3)

Since the above value is negative, volume is maximum for x = 2√(10/3)

Hence volume

= 10 X 2√(10/3)  -  [2√(10/3)]³/4

= 2√(10/3) [10 - 10/3]

= 2√(10/3) X 20/3

= 40√(10/27) cu in

To learn more about Maximization visit

brainly.com/question/14682292

#SPJ4

Complete Question

(Image Attached)

3 0
1 year ago
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