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3 years ago

Write the equation of a line in slope-intercept form from the graph below.

Mathematics

1 answer:

nadezda [96]3 years ago

7 0

Y=x

Slope of 1, y-intercept of (0,0)

y=1x+0...same as the initial answer

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Madison wants to order t-shirts for the university which consists of about 12000 students. she takes a random sample to figure o

earnstyle [38]

Answer:

She should order a total of 5,160 medium sized t-shirts

Step-by-step explanation:

To calculate the number of medium sized t-shirt to be ordered, we shall be using the proportion of students that wanted medium sized t-shirt in her survey to multiply the total number of students that we have.

From the question, we can identify that the probability that a student will like a medium sized t-shirt is simply 129/300

Now we have about 12000 students in the University, using the probability from the survey, the number of medium sized t shirt she should order would be;

129/300 * 12,000 = 129 * 40 = 5,160 medium sized t-shirts

8 0

3 years ago

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

What is the prime factorization of 24

xxMikexx [17]

Answer:The prime factorization is the product of the circled primes. So the

prime factorization of 24 is 24 = 2 · 2 · 2 · 3 = 3.

Step-by-step explanation:

5 0

3 years ago

The solution set of the inequality 2x − y > 2 consists of all the points __ ? the line __ ?

GalinKa [24]

Answer:

b - below; y = 2x − 2

Step-by-step explanation:

2x − y > 2
2x - y + y - 2 > 2 + y - 2
2x - 2 > y
y < 2x - 2

Solution is the points below the line y = 2x - 2

<u>Correct option is:</u>

b - below; y = 2x − 2

7 0

4 years ago

How many different four-digit numbers can be formed with the numbers 7; 4; 5; 1; 2; 9; and 8?

Masja [62]

Answer:

If I'm correct, the answer is 1296.

Step-by-step explanation:

6 x 6 x 6 x 6 = 1296

7 0

3 years ago