The price of 600 grams of candy and 1.5 kg of cookies was 4 dollars and 62 cents. What is the price of one kg of cookies if it i
1 answer:
Answer:
The price of the kilogram of cookies is $ 1.8
Step-by-step explanation:
Call x the price of the kilogram of candy and call y the price of kilogram of cookies
Then we know that
We also know that the price of kilogram of cookies is 1.40 dollars cheaper than the kilogram of candy
So
We solve this equation for x and then we substitute it in the first equation.
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Answer:
Step-by-step explanation:
<u>Part A: the x-intercepts are -3/2 and 5/2 </u>
- 4
- 7x - 15 -----> (2x + 3) (2x - 5)
2x + 3 = 0 --------> 2x = -3 ----------> x =
2x - 5 = 0 ---------> 2x = 5 -----------> x =
<u>Part B: the parabola is a minimum, the vertex is (7/8,-289/16) </u>
f(x)=ax^2+bx+c
if a>0, then the parabola opens up and the vertex is a minimum
a<0 then the parabola opens down and the vertex is a max
- a = 4, it is a minimum
the x value of the vertex in f(x)=ax^2+bx+c= is -b/(2a)
the y value of the vertex is f(-b/(2a))
f(x) = 4 - 7x - 15
a = 4
b = -7
-b/2a=-(7)/(2*4)=7/8
f(7/8) = 4 - 7(7/8) - 15
f(7/8) = 49/16 - 49/8 - 15
f(7/8) = -289/16
- the vertex is (7/8,-289/16)
<u>Part C: </u>
- the vertex is minimum and the graph goes through the x intercepts
- plug in x values to get the y value (EX: choose 0 for x, and you'll get -15 for y, so you would plot the point at (0, -15). you could plug in 1 for x, and get -18 for y, plotting the point (1, -18) )
Answer:
The height equation is:
h(t)=-16t^2 +100t+10
A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.
v(t) = -2*16*t + 100 = 0
t = 100/(2*16) = 3.125s
Now we put this time in our height equation:
h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.
B) The lowest heigth of the potato will be h = 0ft, when the potato hits the ground.
C) the lowest time that works for that function is t = 0s, when the potato is fired by the gun.
D) the maximum time can be finded when the potato hits the ground, afther that point the equation does not work anymore, let's find it,
h(t) = 0 = -16t^2 +100t+10
we can solve it using Bhaskara's equation:
So the two solutions are:
t = 6.3s
t = -0.1s
We need to chose the positive time, as we already discused that the minimum time that works for the equation is t = 0s.
so here the answer is t = 6.3s
E) the domain is 0s ≤ t ≤ 6.3s
The range is 0ft ≤ h ≤ 166.25 ft.