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Solnce55 [7]
2 years ago
8

Which statement accurately explains whether a reflection over the x-axis and a 90° rotation would map figure ACB onto itself? co

unterclockwise ​

Mathematics
1 answer:
miskamm [114]2 years ago
3 0

Answer: Step-by-step explanation: Line AB is horizontal, so reflection across the x-axis maps it to a horizontal line. Then rotation CCW by 90° maps it ...  Which statement accurately explains whether a reflection over the X-axis and a 180° rotation would map figure ACB onto itself?​.

90° counterclockwise. Which statement accurately explains whether a reflection over the x-axis and a 180° rotation would map figure ACB onto itself? Which statement accurately explains whether a reflection over the x-axis and a 90° counterclockwise rotation would map figure ACB onto itself? WILL GIVE IF CORRECT, IF WRONG NO Which statement accurately explains whether a reflection over the x-axis and a 90° counterclockwise rotation would map Answer: 9514 1404 393Answer: No, A″C″B″ is located at A″1, 1, C″4 Which statement accurately explains whether a reflection over the x-axis and a 90° counterclockwise rotation would map figure ACB onto itself? a coordinate Take the point (1,0) that's on the x axis. a 90 degree rotation (counterclockwise of course) makes it be on the y axis instead at (0,1). 90 degrees more is ...

Step-by-step explanation:

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Solve the initial-value problem<br><br> y' = x^4 - \frac{1}{x}y, y(1) = 1.
natta225 [31]

The ODE is linear:

y'=x^4-\dfrac yx

y'+\dfrac yx=x^4

Multiplying both sides by x gives

xy'+y=x^5

Notice that the left side can be condensed as the derivative of a product:

(xy)'=x^5

Integrating both sides with respect to x yields

xy=\dfrac{x^6}6+C

\implies y(x)=\dfrac{x^5}6+\dfrac Cx

Since y(1)=1,

1=\dfrac16+C\implies C=\dfrac56

so that

\boxed{y(x)=\dfrac{x^5}6+\dfrac5{6x}}

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Step-by-step explanation:

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Sum of -6x^2-1 and x+9
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Find the taylor polynomial t3(x) for the function f centered at the number
inysia [295]

Answer:

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

Step-by-step explanation:

We are given that

f(x)=7tan^{-1}(x)

a=1

T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}

Substitute n=3 and a=1

t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}

f(x)=7tan^{-1}(x)

f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}

Where tan^{-1}(1)=\frac{\pi}{4}

f'(x)=\frac{7}{1+x^2}

Using the formula

\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}

f'(1)=\frac{7}{2}

f''(x)=\frac{-14x}{(1+x^2)^2}

f''(1)=-\frac{7}{2}

f''(x)=-14x(x^2+1)^{-2}

f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})

By using the formula

(uv)'=u'v+v'u

f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}

f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}

f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}

Substitute the values

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

7 0
3 years ago
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