What is the vertex form of a parabola with vertex at the origin and the focus at (0,1/8)?

Answers

Answer 1
Answer: Hello,

The parabola of vertex (0,0) has like equation y=ax²
F, the focus is (0,1/(4a)).
Here 1/4a=1/8==>a=2

y=2x²

Answer 2
Answer:

Answer:

f(x)= 2x^2

Step-by-step explanation:


Related Questions

To get from home to work, Felix can either take a bike path through the rectangular park or ride his bike along two sides of the park. How much farther would Felix travel by riding along two sides of the park than he would by taking the path through the park?
Can you solve y=2x-5 and 4x-y=7 with steps?
WILL GIVE POINTS!!! The graph of a function is shown:Which of the following correctly identifies the set of outputs?{(5,-2), (1, -1), (-2, 2), (2,5)}{(-2,5), (-1, 1), (2,-2), (5, 2)}{-2,-1,2,5){-2, 1, 2,5)
The length of x 200 feet​
X^2+6x+c=13+c what is c

mark's computer weighs 35.769 pounds. what is the weight of his computer rounded to the nearest hundredth

Answers

so hundreth is the 2nd number after the decimal point
that is the 6
so to round you take the numer before it so
69
if number is >5 round up

if number is <5 round down

9 is >5 so round up
7

35.77 ppounds
If you round to the nearest hundredth, Marks computer will weigh 35.77 pounds

Find the exact value of cos⁻¹ (cos 5π/7 ). Write your answer in radians in terms of π. If necessary, click on "Undefined."

Answers

Step-by-step explanation:

the cos^-1   and the cos  function cancel each other and you are left with

5pi/7  Radians

To find the midpoint of a segment using the coordinates of its endpoints: Answers: A- Calculate the average of the x-coordinates and the average of the y-coordinates of the endpoints. B- Calculate the differences of the x-coordinates and the differences of the y-coordinates of the endpoints. C-Calculate the differences of the x-coordinates and the differences of the y-coordinates of the endpoints and divide each by 2. D-Calculate the average of the x-coordinates and the average of the y-coordinates of the endpoints and divide each by 2.

Answers

The answer here is A. "Calculate the average of the x-coordinates and the average of the y-coordinates of the endpoints."

The midpoint is basically the point located between two endpoints, where the distance is equal from the midpoint to either of the points. Since this is merely a halfway point, taking the average of the coordinates would yield us the midpoint.

: Ram lived in a small village. When he left for a college in a city, the population of his village was 840. When he came back, the , population had grown by 5%. What was the nopulation he found?​

Answers

Answer:

882

Step-by-step explanation:

840 : 100 x 105 = 882

HELP QUICK Jamie created a list of factors of 32. 32: 1, 2, 4, 6, 8, 16, 32 which number should be removed from the list to make the list accurate? 1 6 16 32

Answers

Answer:

6

Step-by-step explanation:

1: divide 32 by each factor

2:ensure that the answer is a whole number

3: 6, when calculated gives you 5.3(and the threes will go on)

4:You get your answer

Answer:

6

Step-by-step explanation:

The factors listed shows that 1 x 32 = 32, 2x 16= 32, 4 x 8= 32

BUT! When we get to 6 it does not have a number to multiply to that is a whole number. For example, 32/6= 5.333 which is not a whole number. Factors have to be whole numbers and no decimals! So, 6 should be removed from the list to make it accurate!

A crane able can support a maximum load of 15,000 kg. If a bucket has a mass of2,000 kg and gravel has a mass of 1,500 kg for every cubic meter, how many cubicmeters of gravel can be safely lifted by the crane?

Answers

Let

x-------> amount of cubic meters of gravel that can be safely lifted by the crane

we know that

15,000=2,000+1,500*x

Solve for x

15,000=2,000+1,500*x\n1,500*x=15,000-2,000\n1,500*x=13,000\n x=(13,000/1,500)\nx=8.67\ m^(3)

therefore

the answer is

the amount of cubic meters of gravel that can be lifted safely by the crane cannot exceed 8.67\ m^(3)

Well so 15000-2000 leaves 13000kgs of available weight allowed. Then 13000 divided by 1500 we get around 8.67 cubicmeters of gravel to be allowed to be lifted safely.