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mash [69]
3 years ago
14

Find the center and radius of the circle whose diameter has an endpoint at (-3, -4) and the origin.

Mathematics
1 answer:
Zina [86]3 years ago
6 0

Given end points of diameter,

(x1,y1)=(-3,-4)

(x2,y2)=(0,0)

Now,

the equation of circle is,

(x-x1)(x-x2)+(y-y1)(y-y2)=0

or, (x+3)(x-0)+(y+4)(y-0)=0

or, x^2 +3x +y^2 +4y =0

or, x^2 +y^2 +3x + 4y=0

which is in the form of x^2 +y^2 +2gx +2fy + c=0

where,

g=3/2

h=2

c=0

Now,

radius(r) =  \sqrt{ {g}^{2}  +  {f}^{2} - c  } \\  =  \sqrt{ \frac{ {3}^{2} }{ {2}^{2} } +  {2}^{2}  - 0 }  \\  =  \sqrt{ \frac{9}{4}  + 4}  \\  =   \sqrt{ \frac{25}{4} }  \\  =  \frac{5}{2}

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statuscvo [17]

For this case we have that the volume of the figure is composed of the volume of a prism and the volume of a pyramid:

The volume of the prism is given by:

V = A_ {b} * h

Where:

A_ {b}: It is the area of the base

h: It's the height

Substituting:V = 6 * 6 * 6\\V = 216 \ units ^ 3

The volume of the pyramid is given by:

V = \frac {1} {3} * L ^ 2 * h

Where:

L ^ 2:It is the area of the base

h: It's the height

Substituting:

V = \frac {1} {3} * 6 ^ 2 * 5\\V = \frac {1} {3} * 36 * 5\\V = 60units ^ 3

We add and we have:

V = 276 \ units ^ 3

ANswer:

Option D

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3 years ago
What are the coordinates of the image of point A after the segment has been dilated by a scale factor of One-fourth with a cente
Vlada [557]

Answer:

Q=(0.4.-1.7)

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S=(-10.6. 7.3)

Step-by-step explanation:

Just did it and got it wrong and these are the correct answers

6 0
3 years ago
Read 2 more answers
What is the answer to f/3+4=8
Musya8 [376]

Answer:

so it would be f=36

Step-by-step explanation:

subtract 4 to the other side then multiply both sides by 3

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3 years ago
Please do #4 It’s it the one on the bottom
spin [16.1K]

We need to graph this equation:

16x+2y=300

Its solutions are the points through which it graph passes. Since it's a linear equation its graph is a straight line and we only need two of its points to draw it. But before graphing let's re-write the equation. We can substract 16x from both sides:

\begin{gathered} 16x+2y=300 \\ 16x+2y-16x=300-16x \\ 2y=300-16x \end{gathered}

And we divide both sides by 2:

\begin{gathered} \frac{2y}{2}=\frac{300-16x}{2}=\frac{300}{2}-\frac{16x}{2} \\ y=150-8x \end{gathered}

So now with this equation if we pick two random x values we'll get their corresponding y values. This way we'll find two points that are part of the graph which is the line that passes through both. We can begin with x=0:

y=150-8\cdot0=150

So the first point is (0,150). Then we can take x=10 and we get:

y=150-8\cdot10=150-80=70

So the second point is (10,70). Then the graph is the line that passes through points (0,150) and (10,70). In order to represent it

3 0
1 year ago
Rewrite (2x^2+13x+26) / x+4 in the form q(x)+r(x)/b(x) . Then find q(x) and r(x). In the rewritten expression, q(x) is_____and r
likoan [24]

The value of q(x) is 2 x+5

The value of r(x) is 6

Explanation:

The given expression is \frac{2 x^{2}+13 x+26}{x+4}

We need to rewrite the expression in the form of q(x)+\frac{r(x)}{b(x)}

Simplifying the expression, we get,

\frac{2 x^{2}+8 x+5x+26}{x+4}

Separating the fractions, we have,

\frac{2 x^{2}+8 x}{x+4}+\frac{5 x+26}{x+4}

2 x+\frac{5 x+26}{x+4}  -----------(1)

Now, we shall further simplify the term \frac{5 x+26}{x+4} , we get,

\frac{5 x+26}{x+4}=\frac{5 x+20}{x+4}+\frac{6}{x+4}

Common out 5 from the numerator, we have,

\frac{5 x+26}{x+4}=5+\frac{6}{x+4}

Substituting the value \frac{5 x+26}{x+4}=5+\frac{6}{x+4} in the equation(1), we get,

2 x+5+\frac{6}{x+1}

Thus, the expression \frac{2 x^{2}+13 x+26}{x+4}=2 x+5+\frac{6}{x+1} is in the form of q(x)+\frac{r(x)}{b(x)}

Hence, we have,

q(x)=2 x+5

r(x)=6 and

b(x)=x+4

5 0
3 years ago
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