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3 years ago

Choose the correct conic section to fit the equation.

Mathematics

1 answer:

yanalaym [24]3 years ago

5 0

Answer:

Option A) Circle.

Step-by-step explanation:

We are given the equation:

$(x - 3)^2 + (y - 12)^2 = 25$

A general equation of the circle is of the form:

$(x-h)^2 + (y-k)^2 = r^2$

where (h,k) is the center of the circle and r is the radius of circle.

Comparing the two equations, we get,

$(h,k) = (3,12)\\r = \sqrt{25} = 5$

Thus, the given equation is equation of a circle centered at (3,12) and of radius 5 units.

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Describe and correct the error below.

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2 years ago

Which function is even? Check all that apply.

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3 years ago

Please help me solve this problem... I will mark you brainliest

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3 years ago

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melisa1 [442]

Answer:

D. 18.68

Step-by-step explanation:

$27\div \frac{5}{3}-6+53\times (\frac{2}{5})^2$

Applying PEMDAS as order of operations.

Solving the exponents first $[(\frac{2}{5})^2=\frac{4}{25}]$

$=27\div \frac{5}{3}-6+53\times \frac{4}{25}$

Multiplying $[53\times \frac{4}{25}=8.48]$

$=27\div \frac{5}{3}-6+8.48$

Dividing $[27\div \frac{5}{3}=16.2]$

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Adding and subtracting.

$=18.68$

6 0

3 years ago

Help!! Find missing value for each quadratic function

Svetllana [295]

I'll do the first one to get you started. The answer is -3

=================================================

Explanation:

When x = -1, y = 0 as shown in the first column. The second column says that (x,y) = (0,1) and the third column says (x,y) = (1,0)

We'll use these three points to determine the quadratic function that goes through them all.

The general template we'll use is y = ax^2 + bx + c

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

Plug in (x,y) = (0,1). Simplify

y = ax^2 + bx + c

1 = a*0^2 + b*0 + c

1 = 0a + 0b + c

1 = c

c = 1

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

Plug in (x,y) = (-1,0) and c = 1

y = ax^2 + bx + c

0 = a*(-1)^2 + b(-1) + 1

0 = 1a - 1b + 1

a-b+1 = 0

a-b = -1

a = b-1

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

Plug in (x,y) = (1,0), c = 1, and a = b-1

y = ax^2 + bx + c

0 = a(1)^2 + b(1) + 1 ... replace x with 1, y with 0, c with 1

0 = a + b + 1

0 = b-1 + b + 1 ... replace 'a' with b-1

0 = 2b

2b = 0

b = 0/2

b = 0

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

If b = 0, then 'a' is...

a = b-1

a = 0-1

a = -1

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

In summary so far, we found: a = -1, b = 0, c = 1

Therefore, y = ax^2 + bx + c turns into y = -1x^2 + 0x + 1 which simplifies to y = -x^2 + 1

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

The last thing to do is plug x = 2 into this equation and simplify

y = -x^2 + 1

y = -2^2 + 1

y = -4 + 1

y = -3

3 0

3 years ago