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3 years ago

(28 and 29) I really need help

Mathematics

1 answer:

scoundrel [369]3 years ago

5 0

Answer:

28 is 11/60

29 is 7/30

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Find the equation of the tangent line to the curve (a lemniscate)

olya-2409 [2.1K]

Answer:

$m=\frac{9}{13}$ and $b=\frac{40}{13}$

Step-by-step explanation:

The equation of curve is

$2(x^2+y^2)^2=25(x^2-y^2)$

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

$2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})$

$4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})$

The point of tangency is (-3,1). It means the slope of tangent is $\frac{dy}{dx}_{(-3,1)}$ .

Substitute x=-3 and y=1 in the above equation.

$4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})$

$40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})$

$-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}$

$80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240$

$130\frac{dy}{dx}=90$

Divide both sides by 130.

$\frac{dy}{dx}=\frac{9}{13}$

If a line passes through a points $(x_1,y_1)$ with slope m, then the point slope form of the line is

$y-y_1=m(x-x_1)$

The slope of tangent line is $\frac{9}{13}$ and it passes through the point (-3,1). So, the equation of tangent is

$y-1=\frac{9}{13}(x-(-3))$

$y-1=\frac{9}{13}(x)+\frac{27}{13}$

Add 1 on both sides.

$y=\frac{9}{13}(x)+\frac{27}{13}+1$

$y=\frac{9}{13}(x)+\frac{40}{13}$

Therefore, $m=\frac{9}{13}$ and $b=\frac{40}{13}$ .

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2 years ago

Consider the function g defined by g(x) = x(10 – Ax), A > 0. voer g in Without doing any iterations, determine the non-zero f

notka56 [123]

Answer:

The only non-zero fixed point is: x = 9/A.

The Step-by-step explanation:

A fixed point of a function is a points that is mapped to itself by the function; g(x) = x. Therefore, in order to find the fixed point of the given function we need to solve the following equation:

g(x) = x

x(10 - Ax) = x

10x - Ax² = x

10x - x -Ax² = 0

9x - Ax² = 0

Ax² - 9x = 0

The solutions of this second order equation are:

x = 0 and x = 9/A.

Since we are only asked for the non-zero fixed points, the solution is: 9/A.

7 0

3 years ago

Identify the vertex of y = x2 + 4x + 5.

Xelga [282]

ANSWER

The vertex is (-2,1)

EXPLANATION

We want to find the vertex of

$y = {x}^{2} + 4x + 5$

We complete the square to obtain,

$y = {x}^{2} + 4x + {(2})^{2} - {(2})^{2} + 5$

The first three terms forms a perfect square trinomial.

$y = {(x + 2})^{2} - 4 + 5$

The vertex form is

$y = {(x + 2})^{2} + 1$

This equation is in the form;

$y = a{(x -h})^{2} + k$

where (h,k)=(-2,1) is the vertex.

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3 years ago

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