Answer:
15 early tickets were sold and 5 regular tickets were sold.
Step-by-step explanation:
Let x be the number of early bird tickets sold and y be the number of regular tickets sold.
Total number of tickets sold = 20
Thus, we can write the equation,
Cost of early bird ticket = $10
Cost of regular tickets = $15
Total cost = $225
Thus, we can write the equation,
We solve the equation by elimination method by multiplying first equation by 10 and subtracting.
Thus, 15 early tickets were sold and 5 regular tickets were sold.
Answer:
-2.25
Step-by-step explanation:
you gotta divide -4/5 by 2
Answer:
Gradient=¾
Step-by-step explanation:
Write the equation in slope-intercept form y=mx+c
3x-4y=24
-4y=24-3x
y=24/-4-3x/-4
y=-6+¾x
y=¾x-6
The gradient is therefore ¾
The y-intercept of the line occurs at(0,-6)
The x-intercept of the line occurs at(8,0)
Finding the midpoint of the two intercepts
{(x1+x2)/2,(y1+y2)/2}
M={0+8/2,(-6+0/2)}
M={8/2,-6/2}
M=(4,-3)
As you can see, the other information was not required to solve it
Answer:
Step-by-step explanation:
Slope-intercept equation of a line is,
y = mx + b
Here, m = Slope of the line
b = y-intercept
If m = -6 and b = 3,
Equation of the line will be,
y = -6x + 3
Input - output values got the graph of this line,
x -1 0 1 2
y 9 3 -3 -9
Now by plotting these points we can draw the line on graph.
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.
