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3 years ago

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Which one of the following proportions is true for the segments shown in the figure if lines a, b, and c are parallel to each ot

her?

1 answer:

Sever21 [200]3 years ago

8 0

<span>choice B
B. CE/CA = BD/DC </span>

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-1/3(6x-15)+6x>8x-11 solve.

arsen [322]

In this question, you're solving the inequality for x.

Solve:

-1/3(6x - 15) + 6x > 8x - 11

Distribute the -1/3 to the variables in the parenthesis.

-2x + 5 + 6x > 8x - 11

Combine like terms

4x + 5 > 8x - 11

Subtract 8x from both sides

-4x + 5 > - 11

Subtract 5 from both sides

-4x > -16

Divide both sides by -4, while also flipping the inequality, since you're dividing by a negative.

x < 4

Answer:

x < 4

7 0

3 years ago

Express your final answer in simplest form 3/5 + -7/8

frez [133]

Answer:

- 0.5

Step-by-step explanation:

you first change the fractions numbers in to decimal form

so; 2/5 = 0.4

-7/8 = - 0.875

hence; 0.4 + (-0.875)

= - 0.475

in to its simplest form = - 0.5

the answer is - 0.5

5 0

3 years ago

Explain how two amounts of change can be the same but the percent of change can be different

densk [106]

If I understand the question correctly, you can have the same amount of change be two different percentages by having 4 quarters and 100 pennies.
Each quarter is 25%
Each penny is 1%
(am i explaining this right)

5 0

3 years ago

Find the slope of the line that passes through (6, 8) and (9, 10)

valentinak56 [21]

Answer:

m = 2/3

Step-by-step explanation:

The slope of the line is 2/3.

3 0

3 years ago

Read 2 more answers

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