Answer:
29 runs are needed to break the record.
Step-by-step explanation:
He has 32 runs batted in.
If he gets 2 runs batted in at each of the last 13 games, he will have:
32 + 2*13 = 32 + 26 = 58
He will have 58 runs batted in.
He will miss the record by 3.
So the record is 61.
He has 32 and will need 61-32 = 29 runs to break the record.
2x-4=x^2-4
-4=x^2-2x-4 (start moving everything to one side)
x^2-2x=0 (the 4 cancel out)
x(x-2)=0 (factor out an x)
x=0 & x=2 (solved for x)
y=2x-4 (pick one of the equations)
y=2(0)-4 (plug in x)
y=-4
(0,-4)
y=2(2)-4 (plug in the second point)
y=0
(2,0)
The greatest possible value of s is 2. (2,0)
Answer:
(b) 135°
Step-by-step explanation:
The arc has the same measure as the central angle it subtends. This fact is independent of the radius.
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The measure of the central angle is given as 135°, so the measure of the arc is 135°.
The first thing to do in this case is to find the number of inches of wall hanging that the artist needs to weave.
We have then
60-18 = 42 inches
Then, as each day the artist can weave two inches, then the number of days will
n = (42) / (2) = 21 days
answer
it will take 21 days the artist to finish the hanging wall
Let's say "p" people were going to the expedition initially, and the cost for each was "c", now, we know the total cost is 1800, so for "p", folks that'd be 1800/p how much each one cost, namely, how many times "p" goes into 1800.
well, prior to leaving, 15 dropped out, so that leaves us with " p - 15 ", and the cost "c" bumped up to " c + 27 " for each.

![\bf 1800p=1800(p-15)+27[p(p-15)] \\\\\\ 1800p=1800p-27000+27(p^2-15p) \\\\\\ 0=-27000+27(p^2-15p)\implies 0=-27000+27p^2-405p \\\\\\ \textit{now, let's take a common factor of }27 \\\\\\ 0=p^2-15p-1000\implies 0=(p-40)(p+25)\implies p= \begin{cases} \boxed{40}\\ -25 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%201800p%3D1800%28p-15%29%2B27%5Bp%28p-15%29%5D%0A%5C%5C%5C%5C%5C%5C%0A1800p%3D1800p-27000%2B27%28p%5E2-15p%29%0A%5C%5C%5C%5C%5C%5C%0A0%3D-27000%2B27%28p%5E2-15p%29%5Cimplies%200%3D-27000%2B27p%5E2-405p%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20let%27s%20take%20a%20common%20factor%20of%20%7D27%0A%5C%5C%5C%5C%5C%5C%0A0%3Dp%5E2-15p-1000%5Cimplies%200%3D%28p-40%29%28p%2B25%29%5Cimplies%20p%3D%0A%5Cbegin%7Bcases%7D%0A%5Cboxed%7B40%7D%5C%5C%0A-25%0A%5Cend%7Bcases%7D)
well, you can't have a negative value of people... so it has to be 40.
so, 40 folks were initially going, then 15 dropped out, how many went on the expedition? 40 - 15.