The focus of a parabola is located at (4,0), and the directrix is located at x = –4.

1 answer:

4 0

The distance between the focus and directrix is, in this case, 8. Since the directrix is at x= something, the parabola opens sideways, and since the directrix is on the left, to the right. In which case, y^2=p(x), where p is 4 times distance of half of distance between directrix and focus. So, the answer is D. y^2=16x. The 4th answer.

You might be interested in

A population of deer in an area is 2000 and is decreasing at a rate of 15% per year. At this rate what will the deer population

Pepsi [2]

2000 × 15% = 300
300 × 5 = 1500
2000 - 1500 = 500 population after 5 years

4 0

3 years ago

Identify the correct statement for the given figure.

neonofarm [45]

Answer:

Step-by-step explanation:

b/c the figure is right angle

4 0

3 years ago

Read 2 more answers

Tell whether the relation is a function (3,-2)(0,1)(1,0)(-2,-1)(2,-1)

GREYUIT [131]

Have a look at the first elements of each point: {3, 0, 1, -2, 2}. Since these are all different, no one x value has more than one y-value associated with it, and thus (3,-2)(0,1)(1,0)(-2,-1)(2,-1) is a function.

8 0

3 years ago

Read 2 more answers

6) Rotate ARST 90° counterclockwise

allochka39001 [22]

The trick with this is take a piece of paper you can see through and draw the lines of the graph (dont write the numbers) trace the shape and then rotate it the way you need to then line it up with the rest of the graph and you have your answer

8 0

2 years ago

Mr .Baker has baked some muffins. If he packs them in boxes of 4, he will have 3 left over. If he packs them in boxes of 5, he w

babunello [35]

Answer:

$\boxed{43}$

Step-by-step explanation:

I think the easiest way to solve this problem is by brute force: trial and error.

We must find numbers that when divided by

4 leave 3 (4n + 3)
5 leave 3 (5n + 3)
6 leave 1 (6n + 1)

Here is a list of multiples of the integer n that satisfy the three conditions. $\begin{array}{c|ccc}\mathbf{n} & \mathbf{4n+ 3} & \mathbf{5n + 3} & \mathbf{6n + 1}\\1 & 7 & 8 & 7\\2 & 11 & 13 & 13\\3 & 15 & 18 & 19\\4 & 19 & 23 & 25\\5 & 23 & 28 & 31\\6 & 27 & 33 & 37\\7 & 31 & 38 & \mathbf{43}\\8 & 35 & \mathbf{43} & 49\\9 & 39 & 48 & 55\\10 & \mathbf{43} & 53 & 61\\\end{array}\\\text{The only number that is common to all three lists is $\mathbf{43}$.}\\\text{The smallest possible number of muffins Mr. Baker could have baked is $\boxed{\mathbf{43}}$.}$

Check:

$43 \div 4 = 10R3\\43 \div5 = 8R3\\43 \div 6 = 7R1$

OK .

6 0

3 years ago