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3 years ago

Which of these ordered pairs could not be a dilation of (8, -4)? (2, -1) (2, -2) (16, -8) (4, -2)

Mathematics

2 answers:

velikii [3]3 years ago

8 0

The answer is (2,-2). This is because when you set the x value over the y value, it does not simplify to the same ratio, and all of the others do. Take a look at my work...

mr_godi [17]3 years ago

3 0

Answer:

(2, -2)

Step-by-step explanation:

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The length of a rectangle is one more than five times its width. If a perimeter is 38, find the dimensions . Show work .

saul85 [17]

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2 years ago

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77julia77 [94]

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Step-by-step explanation:

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2 years ago

Hazel paid $32 to have 50 reports printed she then paid $57 to have 100 reports printed the relationship is linear what is the e

mihalych1998 [28]

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2 years ago

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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

If sin theta = -0.7660, which of the following represents an approximate value of tan theta, for 180 degrees < theta < 270

Elis [28]

We have that
sin ∅=-0.7660

the sin ∅ is negative
so
∅ belong to the III or IV quadrant
but
180°< ∅ < 270°
hence
∅ belong to the III quadrant
sin ∅=-0.7660
sin² ∅+cos² ∅=1
cos² ∅=1-sin² ∅------> cos² ∅=1-(0.7660)²-----> 0.4132
cos ∅=√0.4132-----> cos ∅=0.6428
the value of cos ∅ is negative-------> III quadrant
cos ∅=-0.6428

tan ∅=sin ∅/ cos ∅----> tan ∅=-0.7660/-0.6428----> tan ∅=1.1917

the answer is
tan ∅ is 1.1917

4 0

3 years ago