All categories

3 years ago

Which of the following solids has no edges?

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

5 0

Answer:

I think it's C a triangular pyramid.

You might be interested in

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

last year 1/5 of the students in your class played a sport this year 10 students join the class five of the new students play a

tino4ka555 [31]

Answer:

Step-by-step explanation:

last year 1/5 of the class played sports, so s = (1/5) c

the next year, s increased by 5, c increased by 10 and the fraction changed to 1/4, so for the next year (s+5) = (1/4)(c+10)

To solve this we can substitute s = (1/5)c from the first equation into the second, so

((1/5)c + 5)=(1/4)(c+10), simplify both sides

(1/5)c + 5 = (1/4)c + 10/4, simplify 10/4 to 5/2

(1/5)c + 5 = (1/4)c + 5/2, multiply both sides by 20 to eliminate fractions (least common multiple of 2, 4 and 5)

4c + 100 = 5c + 50, subtract 4c, subtract 50 from both sides

50 = c, number of students in class last year was 50, this year is 10 more, so this year is 60

5 0

3 years ago

PLEASE HELP WITH QUESTION THANK U

Darya [45]

What question I didn't get it

6 0

3 years ago

What is the following sum?

Nady [450]

Answer:

the answer is c

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Does anyone have answers for the Unit 6-Lesson 9: Rational Expressions and Functions Unit Test? I really need them, and will giv

GalinKa [24]

Answer:

on edge?

Step-by-step explanation:

5 0

3 years ago