The slope would be -2 because the graph is going down and from the y-intercept (0-2), it is going 2 units down and 1 unit right.
Answer:
C. 40.2°
Step-by-step explanation:
Cosine rule (real handy to remember): c² = a² + b² - 2·a·b·cos(γ)
If you don't know this yet, look it up but in short: c, a and b are the lengths of the sides of the triangle, the angle opposite side a is called α, for b it is β and for c it is γ. That's the convention I've always used anyway, you can call them whatever of course. Anyhow:
c² = a² + b² - 2·a·b·cos(γ)
⇒ |AC|² = |AB|²+|BC|²-2·|AB|·|BC|·cos(∠B)
⇒ |AC|²-|AB|²-|BC|² = -2·|AB|·|BC|·cos(∠B)
⇒ ( |AC|²-|AB|²-|BC|² ) / ( -2·|AB|·|BC| ) = cos(∠B)
⇒ ∠B = arccos( ( |AC|²-|AB|²-|BC|² ) / ( -2·|AB|·|BC| ) )
= arccos( ( 11²-16²-16² ) / ( -2·16·16 ) )
= 40.21101958°
≈ 40.2°
Answer:
Nancy paid 2.98 per gallon of gasoline.
Step-by-step explanation:
The price of the gasoline is given by the total amount she paid for the gasoline divided by the amount of gasoline acquired with that money. The calculations for this are shown below:
ratio = total money / total gasoline
ratio = 10.43 / 3.5
ratio = 2.98 money / gallon
Nancy paid 2.98 per gallon of gasoline.
When I see "at what rate", I know this question must come from
pre-Calculus, so I won't feel bad using a little Calculus to solve it.
-- The runner, first-base, and second-base form a right triangle.
The right angle is at first-base.
-- One leg of the triangle is the line from first- to second-base.
It's 90-ft long, and it doesn't change.
-- The other leg of the triangle is the line from the runner to first-base.
Its length is 90-24T. ('T' is the seconds since the runner left home plate.)
-- The hypotenuse of the right triangle is
square root of [ 90² + (90-24T)² ] =
square root of [ 8100 + 8100 - 4320T + 576 T² ] =
square root of [ 576 T² - 4320 T + 16,200 ]
We want to know how fast this distance is changing
when the runner is half-way to first base.
Before we figure out when that will be, we know that since
the question is asking about how fast this quantity is changing,
sooner or later we're going to need its derivative. Let's bite the
bullet and do that now, so we won't have to worry about it.
Derivative of [ 576 T² - 4320 T + 16,200 ] ^ 1/2 =
(1/2) [ 576 T² - 4320 T + 16,200 ] ^ -1/2 times (576T - 4320) .
There it is. Ugly but manageable.
How fast is this quantity changing when the runner is halfway to first-base ?
Well, we need to know when that is ... how many seconds after he leaves
the plate.
Total time it takes him to reach first-base = (90 ft)/(24 ft/sec) = 3.75 sec .
He's halfway there when T = (3.75 / 2) = 1.875 seconds. (Seems fast.)
Now all we have to do is plug in 1.875 wherever we see 'T' in the big derivative,
and we'll know the rate at which that hypotenuse is changing at that time.
Here goes. Take a deep breath:
(1/2) [ 576 T² - 4320 T + 16,200 ] ^ -1/2 times (576T - 4320) =
[ 576 T² - 4320 T + 16,200 ] ^ -1/2 times (1152T - 8640) =
[576(1.875)² - 4320(1.875) + 16,200]^-1/2 times [1152(1.875)-8640] =
[ 2,025 - 8,100 + 16,200 ] ^ -1/2 times [ 2,160 - 8640 ] =
- 6480 / √10,125 = - 64.4 ft/sec.
I have a strong hunch that this answer is absurd, but I'm not going to waste
any more time on it, (especially not for 5 points, if you'll forgive me).
I've outlined a method of analysis and an approach to the solution, and
I believe both of them are reasonable. I'm sure you can take it from there,
and I hope you have better luck with your arithmetic than I've had with mine.
The slope is always the number next to the x. In this case it's 7 or 7/1
