Answer:
Third placer consumed 24 hot dogs.
Winner consumed 48 hot dogs.
Step-by-step explanation:
Lets assume that the third runner up eats x hot dogs.
The total time for the contest is 15x seconds.
Given is that the winner ate (x+24) hot dogs
The winner's rate of eating is 1 hot dog per 7.5 seconds.
And we know the rate = ![\frac{totaltime}{number of hot dogs eaten}](https://tex.z-dn.net/?f=%5Cfrac%7Btotaltime%7D%7Bnumber%20of%20hot%20dogs%20eaten%7D)
We get;
![\frac{15x}{x+24} =7.5](https://tex.z-dn.net/?f=%5Cfrac%7B15x%7D%7Bx%2B24%7D%20%3D7.5)
![(x+24)7.5=15x](https://tex.z-dn.net/?f=%28x%2B24%297.5%3D15x)
![7.5x+180=15x](https://tex.z-dn.net/?f=7.5x%2B180%3D15x)
![7.5x=180](https://tex.z-dn.net/?f=7.5x%3D180)
x = 24 (third placer)
And the winner consumed
hot dogs.
Answer:
Answer would be 9y+36
Step-by-step explanation:
Because if you distribute the 9 inside the parenthesis, you'd get
9*y=9y and 9*4=36
so 9y+36
Hope my answer was helpful to you!
The answer is C because you have three points p to Q, Q to R the two different segments then the whole thing is P to R which is also a segment.
Step-by-step explanation:
Please find the attachment.
We have been given a circle and we are asked to prove that TO is the bisector of angle BTC.
To prove that TO is bisector of angle BTC, we just need to prove that angle BTO is congruent to angle CTO.
We have been given that TB and Tc are tangents to circle O. Since we know that tangents that meet at same point are equal in length.
![TB=TC\text{ ...By two-tangent theorem}](https://tex.z-dn.net/?f=TB%3DTC%5Ctext%7B%20...By%20two-tangent%20theorem%7D)
Since O is the center of our given circle, so OB and OC will be the radii of our given circle.
Since all the points on a circle are equidistant from the center and radius of circle has one one endpoint on the circle and one at the center, so all radii of a circle are congruent.
![OB=OC](https://tex.z-dn.net/?f=OB%3DOC)
We also know that a tangent to a circle is perpendicular to the radius drawn to the point of tangency. As OB and OC are radii and TB and TC are tangents of our given circle,
![m\angle TBO=90^o](https://tex.z-dn.net/?f=m%5Cangle%20TBO%3D90%5Eo)
![m\angle TCO=90^o](https://tex.z-dn.net/?f=m%5Cangle%20TCO%3D90%5Eo)
We can see in our triangles TBO and TCO that,
![TB=TC](https://tex.z-dn.net/?f=TB%3DTC)
![m\angle TBO=m\angle TCO](https://tex.z-dn.net/?f=m%5Cangle%20TBO%3Dm%5Cangle%20TCO)
![OB=OC](https://tex.z-dn.net/?f=OB%3DOC)
Therefore, by SAS congruence
.
So by corresponding parts of congruent triangles are congruent
, therefore, TO is the bisector of
.