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Alexeev081 [22]
1 year ago
5

A circle has a circumference of 33 inches. What is the area of the circle?

Mathematics
1 answer:
UkoKoshka [18]1 year ago
6 0

Answer: Circle

Solve for area

A≈86.66in²

C Circumference

Using the formulas

A=πr2

C=2πr

Solving forA

A=C2

4π=332

4·π≈86.65987in²

Step-by-step explanation:

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Determine the formula for the nth term of the sequence:<br>-2,1,7,25,79,...​
rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

x_n = a n^4 + b n^3 + c n^2 + d n + e

From the given known values of the sequence, we have

\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}

Solving the system yields coefficients

a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4

so that the n-th term in the sequence might be

\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}

Then the next few terms in the sequence could very well be

\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of \{x_n\} eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of \{x_n\} by \Delta^{k}\{x_n\}. Then

• 1st-order differences:

\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}

• 2nd-order differences:

\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}

• 3rd-order differences:

\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}

• 4th-order differences:

\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}

From here I made the assumption that \Delta^4\{x_n\} is the constant sequence {15, 15, 15, …}. This implies \Delta^3\{x_n\} forms an arithmetic/linear sequence, which implies \Delta^2\{x_n\} forms a quadratic sequence, and so on up \{x_n\} forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

5 0
2 years ago
A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.
noname [10]

Answer:

Width = 4ft

Height = 4ft

Length = 8ft

Step-by-step explanation:

Given

Volume = 128ft^3

L = 2W

Base\ Cost = \$9/ft^2

Sides\ Cost = \$6/ft^2

Required

The dimension that minimizes the cost

The volume is:

Volume = LWH

This gives:

128 = LWH

Substitute L = 2W

128 = 2W * WH

128 = 2W^2H

Make H the subject

H = \frac{128}{2W^2}

H = \frac{64}{W^2}

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

A = LW + 2(WH + LH)

The cost is:

Cost = 9 * LW + 6 * 2(WH + LH)

Cost = 9 * LW + 12(WH + LH)

Cost = 9 * LW + 12H(W + L)

Substitute: H = \frac{64}{W^2} and L = 2W

Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)

Cost =18W^2 +  \frac{768}{W^2}*3W

Cost =18W^2 +  \frac{2304}{W}

To minimize the cost, we differentiate

C' =2*18W +  -1 * 2304W^{-2}

Then set to 0

2*18W +  -1 * 2304W^{-2} =0

36W - 2304W^{-2} =0

Rewrite as:

36W = 2304W^{-2}

Divide both sides by W

36 = 2304W^{-3}

Rewrite as:

36 = \frac{2304}{W^3}

Solve for W^3

W^3 = \frac{2304}{36}

W^3 = 64

Take cube roots

W = 4

Recall that:

L = 2W

L = 2 * 4

L = 8

H = \frac{64}{W^2}

H = \frac{64}{4^2}

H = \frac{64}{16}

H = 4

Hence, the dimension that minimizes the cost is:

Width = 4ft

Height = 4ft

Length = 8ft

8 0
2 years ago
Dy/dx = y/x , y(1) = −2
s2008m [1.1K]

Start with

\dfrac{dy}{dx}=\dfrac{y}{x}

Separate the variables:

\dfrac{dy}{y} = \dfrac{dx}{x}

Integrate both parts:

\displaystyle \int \dfrac{dy}{y} = \int\dfrac{dx}{x}

Which implies

\log(y) = \log(x)+c

Solving for y:

y = e^{\log(x)+c} = e^{\log(x)}e^c=xe^c

Since e^c is itself a constant, let's rename it c_1.

Fix the additive constant imposing the condition:

y(1) = c_1\cdot 1 = -2\iff c_1=-2

So, the solution is

y(x) = -2x

5 0
3 years ago
What percentage is 1100
kherson [118]
The answer is 100% percent
6 0
3 years ago
Read 2 more answers
I NEED HELP WILL GIVE BRAINLIEST!! PLEASE HELP ME ASAP!!!!
GenaCL600 [577]

Answer:

A. The length of the second leg is 8.5 inches

B. The length of the three-dimensional diagonal is 9.9 inches

Step-by-step explanation:

Let us revise the relation between the hypotenuse and the two legs of a right triangle

(hypotenuse)² = (vertical leg)² + (horizontal leg)²

∵ The length of the rectangular box = 8 inches

∵ The width of the rectangular box = 3 inches

∵ The height of the rectangular box = 5 inches

∵ Length and width are perpendicular to each other

∴ The Δ whose legs are 3 and 8 is a right triangle

In the right Δ whose legs are 3 and 8

∵ (hypotenuse)² = (3)² + (8)²

∴ (hypotenuse)² = 9 + 64

∴ (hypotenuse)² = 73

- Take √  for both sides

∴ hypotenuse = \sqrt{73} = 8.544003745

- Round it to the nearest tenth of one inch

∴ hypotenuse = 8.5 inches

A.

The 3-dimensional diagonal is the hypotenuse of a right triangle whose legs are the vertical edge and the hypotenuse of the right triangle whose legs are 3 and 8

∵ The hypotenuse of the right triangle whose legs are 3 and

   8 is 8.5 inches

∴ The length of the second leg is 8.5 inches

B.

In the right triangle whose hypotenuse is the 3-dimensional diagonal and legs are the vertical edge , the hypotenuse of the right triangle whose legs are 3 and 8

∵ (3-dimensional diagonal)² = (5)² + (73)²

∴ (3-dimensional diagonal)² = 25 + 73

∴ (3-dimensional diagonal)² = 98

- Take √ for both sides

∴ 3-dimensional diagonal = \sqrt{98} = 9.899494937

- Round it to the nearest tenth of an inch

∴ 3-dimensional diagonal = 9.9 inches

∴ The length of the three-dimensional diagonal is 9.9 inches

<em>V.I.N: you can find the length of the  three-dimensional diagonal by using this rule → </em>d=\sqrt{l^{2}+w^{2}+h^{2}}<em> </em>

3 0
3 years ago
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