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2 years ago

12

Find an ordered pair (x, y) that is a solution to the equation. -x+4y= 2

1 answer:

brilliants [131]2 years ago

8 0

Answer:

x=0 y=0.5

Step-by-step explanation:

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Not sure how to complete #14

pashok25 [27]

14. Add the line times the outside number
X(X+4)= (2√3)^2
X^2+4X= 4√9
X^2+4X= 4times 3
X^2+4X=12
X^2+4X-12=0
(X+6)(X-2)=0
X=-6 X=2
Answe can't be negative so it's X=2

3 0

2 years ago

3x - 6y = 27<br><br> 3x + 4y = 17

11111nata11111 [884]

Answer: {x,y} = {27/4,-47/8}

3 0

2 years ago

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Using ratio pleases help i´m stuck i need #5 for one side 1-5 for the other

Fudgin [204]

How did you put multiple pictures on this app? i can’t figure how to

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2 years ago

Graph (-4,5), (2,3), (-3.0). (-4,-5). (-5. 2). and (-5.3) and connect the points to form a polygon. How many sides does the poly

Keith_Richards [23]

Answer:

6.

Step-by-step explanation:

Source: Desmos.

When the points are plotted, the shape is a polygon. The polygon has 6 sides.

Picture represents the polygon shape, and its sides

8 0

2 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

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

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago