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3 years ago

Using the following equation, find the center and radius of the circle: x^2 + y^2 +2x − 4y − 20 = 0

1 answer:

6 0

You have to complete the squares on both the x terms and the y terms in order to solve this. Move the 20 over to the other side so it's negative. Group the x terms together and complete the square to get (x^2+2x+1) and then do the same with the y terms: (y^2-4y+4). You have to add 1 and 4 to other side with the 20 to get a 25. Then create 2 perfect square binomials within each x and y value to get the vertex coordinates: (x+1)^2 + (y-2)^2 = 25. This tells us that the vertex is located at (-1, 2) and the radius is the square root of 25 which is 5.So the answer is the first choice above.

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6+x3=1

Vadim26 [7]

Answer:

$6 + \frac{x}{3} = 1 \\ \frac{x}{3} = 1 - 6 \\ \frac{x}{3} = - 5 \\ x = 3 \times ( - 5) \\ \boxed{x = - 15}✓$

★ 6+x/3=1

What is the solution to the equation?

x=−21
x=−5
x = 21
<u>x=−15</u>

<h3><u>x= -15</u> is the right answer.</h3>

6 0

3 years ago

Ted is asked to multiply 10 to the second power x 18.72. How Should he move the decimal point

sladkih [1.3K]

Move it two times to the right

7 0

3 years ago

Can anyone tell me the standard form of this parabola equation?

irina [24]

Answer:

$x=4y^2-24y+30$

$(x-4.5)^2+(y-2.5)^2=23$

Step-by-step explanation:

Standard form of a sideways parabola: $x=ay^2+by+c$

Given equation:

$4y^2-x-24y+30=0$

Add x to both sides:

$4y^2-24y+30=x$

$\implies x=4y^2-24y+30$

--------------------------------------------------------------------------------------------

<u>Standard form of circle equation</u>

$(x-a)^2+(y-b)^2=r^2$

(where (a,b) is the center and r is the radius)

Given equation:

$2x^2+2y^2-18x-10y+7=0$

Group like terms:

$2x^2-18x+2y^2-10y+7=0$

Divide by 2:

$x^2-9x+y^2-5y+3.5=0$

Factor by completing the square for each variable:

$(x-4.5)^2-20.25+(y-2.5)^2-6.25+3.5=0$

Rearrange into standard form:

$(x-4.5)^2+(y-2.5)^2=23$

Therefore, the circle has a center at (4.5, 2.5) and a radius of √23

7 0

2 years ago

Find dy/dx when y = Ln (sinh 2x). A. 2 cosh 2x B. 2 sech 2x C. 2 coth 2x D. 2 csch 2x

givi [52]

The function given is a composite function. Let's work from the outside in:
$y=(ln(sinh(2x)))$

First step:
$\frac{d}{dx}[y]= \frac{d}{dx}[ln(sinh(2x))]$

Now, let's work it out:
$\frac{dy}{dx} = \frac{1}{sinh(2x) } * \frac{d}{dx}[sinh(2x)]$

Next step:
$\frac{dy}{dx} = \frac{1}{sinh(2x) } * cosh(2x) * \frac{d}{dx}[2x]$

Next step:
$\frac{dy}{dx} = \frac{1}{sinh(2x) } * cosh(2x) *2$

Simplify:
$\frac{dy}{dx} = \frac{2cosh(2x)}{sinh(2x) }$

Simplify further:
$\frac{dy}{dx} = 2 (\frac{cosh(2x)}{sinh(2x)})$

Remember that:
${\frac{cosh(x)}{sinh(x)} = coth(x)$

So, your final answer is:
$\boxed{ \frac{dy}{dx} = 2coth(2x) }$

So, your answer is C. Hope I could help you!

8 0

3 years ago

Which dot plot matches the tally chart below?

Zepler [3.9K]

Answer:

option 1

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers