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faltersainse [42]
3 years ago
9

The system of equations is graphed on the coordinate plane. y=−12x−1y=14x−4 A graph with a blue line passing through coordinates

(0, -4) and (4, -3) and a red line passing through coordinates (0, -1) and (4, -3). Enter the coordinates of the solution to the system of equations in the boxes. ( , ​ )

Mathematics
1 answer:
Anika [276]3 years ago
8 0
We have that

 case 1)
<span>system of equations is
</span><span>y=−12x−1
y=14x−4

using a graph tool
see the attached figure

the solution of the system  is the point (0.115,-2.385)

case 2)
</span>system of equations is
<span>blue line passing through coordinates A (0, -4) and B (4, -3)
</span><span>red line passing through coordinates   C(0, -1) and  D (4, -3)

</span>using a graph tool
see the attached figure

the solution of the system  is the point (4,-3)<span>

</span>

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Multiplying by 2 to get the common denominator.

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3 years ago
I don’t know what to do
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3 years ago
Find the taylor polynomial t3(x) for the function f centered at the number
inysia [295]

Answer:

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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)

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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}

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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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ivanzaharov [21]
2.27 i just divided 12 to 327 then too the the answer that i had gotten
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3 years ago
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