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2 years ago

A club is selling hats and jackets as a fundraiser.

Mathematics

1 answer:

Neporo4naja [7]2 years ago

6 0

Step-by-step explanation:

150√389/250x8/234

Then the answer gou get add it

15056 ""%90=233.567

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1. Where will the line y = -1/2 -3 cross the y-axis?

soldier1979 [14.2K]

1) The y- axis is where x = 0, so the line "y = -1/2x - 3" will cross the y-axis when x = 0. You plug in "x = 0" into the equation, and get y = 0 -3, so y = -3. That means the line " y = -1/2x - 3" will cross the y-axis ar (0, -3)

2) If there are 2 parallel lines, then there is no solution, because the definition of parallel lines is, " 2 lines in the same plane, that never meet." That makes the solution inconsistent.

7 0

2 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

2 years ago

The proof that AMNS AQNS is shown.

FrozenT [24]

Answer:

The answer is "MS and QS".

Step-by-step explanation:

Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.

As NR and MQ bisect each other at S

⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ

In ΔMNS and ΔQNS

MN=QN (∵ MNQ is isosceles triangle)

∠NMS=∠NQS (∵ MNQ is isosceles triangle)

MS=SQ (Given)

By SAS rule, ΔMNS ≅ ΔQNS.

Hence, segments MS and SQ are therefore congruent by the definition of bisector.

The correct option is MS and QS

3 0

2 years ago

PLEASE ANSWER THIS: look at the pic for question. Thanks!!!

Nadya [2.5K]

Answer:

$\text{D. }b^2-4ac>0$

Step-by-step explanation:

The equation $b^2-4ac$ represents the discriminant of a quadratic. It is the part taken from under the radical in the quadratic formula.

For any quadratic:

If the discriminant is positive, or greater than 0, the quadratic has two solutions

If the discriminant is equal to 0, the quadratic has one distinct real solution (the solution is repeated).

If the discriminant is negative, or less than 0, the quadratic has zero solutions

In the graph, we see that the equation intersects the x-axis at two distinct points. Therefore, the quadratic has two solutions and the discriminant must be positive. Thus, we have $b^2-4ac>0$ .

4 0

2 years ago

What are the sum of the two pink angles?

jenyasd209 [6]

Answer:

45 degrees

Step-by-step explanation:

5 0

3 years ago