An earthworm of length 15cm is crawling along at 2cm/s. an ant overtakes the worm in 5 seconds. how fast is the ant walking?

Assume we have a number line, where the tail of the worm is at x=0. This means that the head of the worm is at x=15.

Also, we know that the coordinate of the head of the worm follows this equation:

Where is the time in seconds. In fact, we know that the worm gains 2 centimeters per second.

Now, assume that the ant also starts at x=0, and walks with a certain rate of centimeters per second. This means that the equation for the position of the ant is

Now, we know that after 5 seconds the ant is in the same position as the head of the worm, which means

We know the equation for both expressions, so let's subtitute them in:

So, the ant walks at a rate of 5cm per second.

i dont know the answer but here's a picture of a dog

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)