Sal was tired of waiting for amir to show up at their meeting place, so he started running at an average speed of 4.5 mph without him. ten minutes later, amir still had 0.25 of a mile to go before he reached the meeting place, but sal was 0.75 of a mile beyond it. if amir is running at a faster pace, 6.7 mph, to the nearest hundredth of an hour, after how much time will amir catch up to sal?

Hence, after 0.45 hour Amir will catch up to Sal.

Step-by-step explanation:

Let A(t) and S(t) denote the function that represent the  distance of that Sal and Amir covers over a period of time 't'.

We have to find the value of t such that:

S(t)=A(t)

As it is given that:

Amir is running at a faster pace, 6.7 mph, Amir still had 0.25 of a mile to go before he reached the meeting place.

so,

A(t)= -0.25+6.7t

Also,

Sal is running at an average speed of 4.5 mph, but Sal was 0.75 of a mile beyond it.

This means that:

S(t)=0.75+4.5t

Also now,

A(t)=S(t)

⇒  -0.25+6.7t=0.75+4.5t

⇒ 6.7t-4.5t=0.75+0.25

⇒ 2.2t=1

⇒ t=1/2.2

⇒ t=5/11

⇒ t≈0.45 hr.

Hence, after 0.45 hour Amir will catch up to Sal.

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.

after 1 year: 221 deer

step-by-step explanation:

30% of 170 is 51. 170+51=221