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2 years ago

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

Mathematics

1 answer:

UkoKoshka [18]2 years 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: -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

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

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

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

3 0

3 years ago