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BabaBlast [244]
3 years ago
6

The diameter of an ice skating rink is 44 feet. In terms of π, what is the area of the rink to the nearest square yard? (9 ft2 =

1 yd2)
Mathematics
2 answers:
Dmitry [639]3 years ago
8 0

Answer:

B. 54π yd^2

Step-by-step explanation:

On god it is right trust mee!    

Over [174]3 years ago
7 0

Answer: Our required area is 54π sq. yd.

Step-by-step explanation:

Since we have given that

Diameter of an ice skating = 44 feet

Radius of an ice skating = 22 feet

as we know the formula for area :

Area would be

\pi r^2\\\\=\pi \times 22^2\\\\=484\pi\ ft^2

As we know that

9 sq ft = 1 sq yd

So, it becomes,

\dfrac{484\pi}{9}=53.77\pi\ yd^2\approx 54\pi\ yd^2

Hence, our required area is 54π sq. yd.

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-2+n=-8(1-8n)+6 solve for n.
blsea [12.9K]

-2 + n = -8(1 - 8n) + 6

Distribute the -8

-2 + n = -8 + 64n + 6

Combine like terms.

-2 + n = 64n - 2

Add 2 to both sides and subtract n from both sides.

0 = 63n

Divide both sides by 63

0 = n

7 0
3 years ago
A point P is moving along the curve whose equation is y = \sqrt x . Suppose that x is increasing at the rate of 4 units/s when x
Mice21 [21]

Answer:3 units/s

Step-by-step explanation:

Given

y=\sqrt{x}

Point P lie on this curve so any general point on curve can be written as (x,\sqrt{x})

and \frac{\mathrm{d} x}{\mathrm{d} t}=4 units/s

Distance between Point P and (2,0)

P=\sqrt{(x-2)^2+(\sqrt{x}-0)^2}

P at x=3 P=2

rate at which distance is changing is

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{\mathrm{d} \sqrt{(x-2)^2+(\sqrt{x}-0)^2}}{\mathrm{d} t}

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2x-3}{\sqrt{(x-2)^2+(\sqrt{x}-0)^2}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2\times 3-3}{2\times 2}\times 4=3 units/s

8 0
3 years ago
Using the equation 2 + 3/41 = 5, what does the value of I equal?
Wewaii [24]

Answer:

i did't see the value I in the equation

Step-by-step explanation:

8 0
3 years ago
Help please <br> x=6<br> x= square root 80<br> x=12<br> x= square root 164
AURORKA [14]

Answer:

x = 12

Step-by-step explanation:

8 0
3 years ago
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

6 0
3 years ago
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