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3 years ago

510

Mathematics

1 answer:

Hoochie [10]3 years ago

6 0

Answer:

$574.41

Step-by-step explanation:

multiply 30 by dollar to pesos

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Find the equation of the tangent line. y=(x^2+x-2)^2 at (-1,4)

Keith_Richards [23]

Compute the derivative of y = (x² + x - 2)² using the chain rule:

dy/dx = 2 (x² + x - 2) d/dx [x² + x - 2]

dy/dx = 2 (x² + x - 2) (2x + 1)

Evaluate the derivative at x = -1 :

dy/dx (-1) = 2 ((-1)² + (-1) - 2) (2 (-1) + 1) = 4

This is the slope of the tangent line to the function at (-1, 4).

Use the point-slope formula to get the equation for the tangent line:

y - 4 = 4 (x - (-1)) → y = 4x + 8

6 0

3 years ago

Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

3 years ago

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, -1).

ira [324]

Line:
3x+y=7
y=-3x+7

Slope of Perpendicular Line: 1/3x

For the perpendicular line to contain point (6,-1) the y-intercept would be (0,1), thus the equation of the line would be y=1/3x+1

y=1/3x+1

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3 years ago

What are the medians and modes??

DerKrebs [107]

Answer:

The median is 17 and the mode is 19

Step-by-step explanation:

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Zigmanuir [339]

Step-by-step explanation:

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3 years ago