24. Solve (x-3⅓)² = 5⁴/9

If x^2 + y^2 = 25, find the equation of the line tangent to the circle when x = -4

bekas [8.4K]

Answer:

In the point (-4,3) tangent is : y=4/3x+25/3 i.e, y=1.33x+8.33

In the point (-4,-3) tangent is: y=-4/3x-25/3 i.e, y=-1.33x-8.33

Step-by-step explanation:

The center of the circle is in (0,0), because x^2+y^2=(x-0)^2+(y-0)^2=25. And the radius is 5, because 5^2=25.

We have first center: (x1,y1)=(0,0)

The point of the tangent is in x=-4, so x^2+y^2=25 is 16+y^2=25,i.e y=3 and y=-3. Like you can see we have 2 tangents to the circle when x=-4. the point (-4,3) and (-4,-3).

We will do for the first point:

The point of tangent is (x2,y2)=(-4,3)

The gradient of the radius

mr=(y2-y1)/(x2-x1)=(3-0)/(-4-0)=-3/4

The gradient of the tangent mt is:

mr*mt=-1 (they are ortogon)

mt=-1/mr=-1/(-3/4)=4/3

the equation of the line trought one point is:

y-y2=k(x-x2), where k is mt,i.e. k=4/3

y-3=4/3(x-(-4))

y=4/3x+16/3+3

y=4/3x+25/3

y=1.33x+8.33

Now for the 2nd point (-4,-3),

mr=-3/-4=3/4

mt=-4/3

y+3=-4/3(x-(-4))

y=-4/3x-16/3-3

y=-4/3x-25/3

y=-1.33x-8.33

7 0

3 years ago

israel started to solve a radical equation in this way: square root of x plus 6 − 4 = x square root of x plus 6 − 4 4 = x 4 squa

agasfer [191]

Lets solve our radical equation $\sqrt{x+6}-4=x$ step by step.

Step 1 add 4 to both sides of the equation:

$\sqrt{x+6} -4+4=x +4$

$\sqrt{x} +6=x+4$

Step 2 square both sides of the equation:

$\sqrt{x+6} =x+4$

$\sqrt{x+6}^2 =(x+4)^2$

$x+6=(x+4)^2$

Step 3 expand the binomial in the right hand side:

$x+6=x^2+8x+16$

Step 4 simplify the expression:

$0=x^2+8x-x+16-6$

$x^2+7x+10=0$

Step 5 factor the expression:

$(x+2)(x+5)=0$

Step 6 solve for each factor:

$x+2=0$ or $x+5=0$

$x=-2$ or $x=-5$

Now we are going to check both solutions in the original equation to prove if they are valid:

For $x=-2$

$\sqrt{x+6}-4=x$

$\sqrt{-2+6}-4=-2$

$\sqrt{4}-4=-2$

$2-4=-2$

$-2=-2$

The solution $x=2$ is a valid solution of the rational equation $\sqrt{x+6}-4=x$ .

For $x=-5$

$\sqrt{x+6}-4=x$

$\sqrt{-5+6}-4=-5$

$\sqrt{1}-4=-5$

$1-4=-5$

$-3\neq -5$

Since -3 is not equal to -5, the solution $x=-5$ is not a valid solution of the rational equation $\sqrt{x+6}-4=x$ ; therefore, $x=-5$ is an extraneous solution of the equation.

We can conclude that even all the algebraic procedures of Israel are correct, he did not check for extraneous solutions.

An extraneous solution of an equation is the solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation. Is worth pointing out that extraneous solutions are particularly frequent in rational equation.

8 0

4 years ago

Help plz worth 20 points

svetoff [14.1K]

Answer:

y=1/4x-4

Step-by-step explanation:

4 0

3 years ago

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Which of the following is the value of e^2 correct to 3 decimal places

Tcecarenko [31]