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3 years ago

GIVING BRAINLIEST TO PERSON WHO GIVES CORRECT ANSWER TO THIS CONIC SECTIONS QUESTION

1 answer:

4 0

Circle 1's center is at x=-4 and Circle 2's center is at x=5. To translate Circle 1 to Circle 2, we need to shift the circle to the right 9 places:

5 - (-4) = +9

We do the same thing for the y value. We are going from y=-2 to y=3.

3 - (-2) = +5

So the first blank is (9, 5)

Next, the scale factor refers to the factor at which the circle is stretched or shrunk. We find it by comparing the radii. Going from 3 cm to 6 cm, the scale factor is:

6/3 = 2.

The scale factor is 2.

You might be interested in

Angles that measure between 0 and 90 degrees are called_?

azamat

Answer:

Acute angles.

Step-by-step explanation:

Acute angles are from 0-90 degrees.

Hope this helps :)

7 0

3 years ago

Read 2 more answers

What’s A cylinder with a radius of 8yd and a 12 height of yd

steposvetlana [31]

Answer:

192π yd²

Step-by-step explanation:

Surface area of a cylinder = 2πrh

Given, r = 8yd , h = 12yd

S. A = 2 × π × 8 × 12

S. A = 192π yd²

I hope this was helpful, please mark as brainliest

7 0

3 years ago

A = πηγ2<br> What do you divide by?

defon

Answer:

76.20+q; 76.60Es

Step-by-step explanation:

5 0

3 years ago

A drawing has a scale 1/4 in: 12ft. How long is 8ft?

allochka39001 [22]

If you use a proportion, you'll get:

$\frac{0.25}{12} = \frac{x}{8}$

Then, you use cross products:

$12x=0.25(8) \\ 12x=2$

Then, you algebraically solve x:

$12x=2 \\ \frac{12x}{12} = \frac{2}{12} \\ x= \frac{5}{3}$

The answer is 5/3 inches , or 1 5/13 inches, or 1.6 repeating.

4 0

3 years ago

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is $\bold{W=-3e^{7x}}$

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

$W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|$

Thus replacing the functions of the exercise we get:

$W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|$

Working with the determinant we get

$W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}$

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0

3 years ago