a. Answer: D: (∞, ∞)
R: (-∞, ∞)
<u>Step-by-step explanation:</u>
Theoretical domain is the domain of the equation (without an understanding of what the x-variable represents).
Theoretical range is the range of the equation given the domain.
c(p) = 25p
There are no restrictions on the p so the theoretical domain is All Real Numbers.
Multiplying 25 by All Real Numbers results in the range being All Real Numbers.
a) D: (∞, ∞)
R: (-∞, ∞)
*********************************************************************************
b. Answer: D: (0, 200)
R: (0, 5000)
<u>Step-by-step explanation:</u>
Practical domain is the domain of the equation WITH an understanding of what the x-variable represents.
Practical range is the range of the equation given the practical values of the domain.
The problem states that p represents the number of cups. Since we can't have a negative amount of cups, p ≥ 0. The problem also states that Bonnie will purchase a maximum of 200 cups. So, 0 ≤ p ≤ 200
The range is 25p → (25)0 ≤ (25)p ≤ (25)200
→ 0 ≤ 25p ≤ 5000
b) D: (0, 200)
R: (0, 5000)
Answer:
6
Step-by-step explanation:
The mode is the number that appears most in the data set.
So your answer would be 6 because it appears 3 times.
Answer:
y=t−1+ce
−t
where t=tanx.
Given, cos
2
x
dx
dy
+y=tanx
⇒
dx
dy
+ysec
2
x=tanxsec
2
x ....(1)
Here P=sec
2
x⇒∫PdP=∫sec
2
xdx=tanx
∴I.F.=e
tanx
Multiplying (1) by I.F. we get
e
tanx
dx
dy
+e
tanx
ysec
2
x=e
tanx
tanxsec
2
x
Integrating both sides, we get
ye
tanx
=∫e
tanx
.tanxsec
2
xdx
Put tanx=t⇒sec
2
xdx=dt
∴ye
t
=∫te
t
dt=e
t
(t−1)+c
⇒y=t−1+ce
−t
where t=tanx
Answer:
-22
Step-by-step explanation:
The blue dot on the number line is 4 units below -10. The number -16 is also labeled, but it is instead just 2 units below -10. If -16 is 2 units below -10, then 2 units is equal to a distance of 6, which means 1 unit is equal to a distance of 3. Because the blue dot is 4 units below -10, the blue dot is 12 less than -10, meaning that the blue dot is at -22.
Answer:
3rd one
Step-by-step explanation:
any 2 squares are congruent since all four sides of any square are congruent