Given a circle with a diameter of 8, what is the circumference?

umka2103 [35]

Short answer = false
Remark
If it is a right triangle, it will obey the Pythagorean Equation. a^2 + b^2 = c^2. Let's see if it does.

Equation
a^2 + b^2= c^2
c must be the longest line. (Hypotenuse)

Givens
a = 24
b = 17
c = 40

Sub and Solve
a^2 + b^2 = c^2
17^2 + 24^2 = c^2 [We'll see if this comes out to 40]

289 + 576 = c^2
865 = c^2 which is no where near what 40^2 equals. 40^2 = 1600

Short answer False

Square root of 865 = (865)^1/2 = 29.4

5 0

3 years ago

Power Series Differential equation

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$

4 0

3 years ago

Find the area of the figure.

Nady [450]

Answer:

Explanation:

So you want to start by finding the area of the big square.

We know the width to the big square is 60cm. The length to the big square is

19

+

24

+

13

=

56

Big square

A

r

e

a

=

L

⋅

W

=

56

⋅

60

=

3360

Now let’s find the little square inside the bigger square. We see from the picture that the width is 34 and the length is 24cm. Using the same formula, we can find the area of the little square.

Little square

A

r

e

a

=

L

⋅

W

=

24

⋅

34

=

816

The area enclosed is the area of the big square minus the area of the little square.

3360

−

816

=

2544

So the area enclosed is

2544

c

m

2

Step-by-step explanation:

4 0

2 years ago

Sooo ummm help... 15-p=p-3

bearhunter [10]

Answer:

p = 9

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A company needs $5,400,000 in 14 years in order to expand their factory. How much should the company invest each week if the inv

ludmilkaskok [199]

Answer: the company should invest $12191 each week

Step-by-step explanation:

The amount that the company needs is $5,400,000

We would apply the periodic interest rate formula which is expressed as

P = a/[{(1+r)^n]-1}/{r(1+r)^n}]

Where

P represents the weekly payments.

a represents the amount that the company needs

r represents the rate.

n represents number of weekly payments. Therefore

a = 5,400000

There are 52 weeks in a year

r = 0.079/52 = 0.0015

n = 52 × 14 = 728

Therefore,

P = 5400000/[{(1+0.0015)^728]-1}/{0.0015(1+0.0015)^728}]

5400000/[{(1.0015)^728]-1}/{0.0015(1.0015)^728}]

P = 5400000/{2.98 -1}/[0.0015(2.98)]

P = 5400000/(1.98/0.00447)

P = 5400000/442.95

P = $12191

3 0

3 years ago