Answer:
69
Step-by-step explanation:
Answer:
The proof is detailed below.
Step-by-step explanation:
We will first prove that if H(x) is a differentiable function in [a,b] such that H'(x)=0 for all x∈[a, b] then H is constant. For this, take, x,y∈[a, b] with x<y. By the Mean Value Theorem, there exists some c∈(x,y) such that H(y)-H(x)=H'(c)(x-y). But H'(c)=0, thus H(y)-H(x)=0, that is, H(x)=H(y). Then H is a constant function, as it takes the same value in any two different points x,y.
Now for this exercise, consider H(x)=F(x)-G(x). Using differentiation rules, we have that H'(x)=(F-G)(x)'=F'(x)-G'(x)=0. Applying the previous result, F-G is a constant function, that is, there exists some constant C such that (F-G)(x)=C.
Garden one and two are both unknown, so I am choosing to call garden 2 X and then label garden 1 with comparisons to X.
It is tempting to list:
Garden Two = X
Garden One = X + 9
BUT there is an easier way. We are told that when 3 bushes are taken from garden 2 (x-3) and put in garden 1 (x + 12) then garden one has 1.5 times more than garden two.
Set it up like this:
1.5 ( x - 3) = x + 12 (because 1.5 times garden 2 will give us garden 1)
1.5x - 4.5 = x + 12 (Distribute)
.5x = 16.5 (Use subtraction to move variables to the right and other numb left)
x = 33 for Garden 2
33 + 9 for Garden 1 = 42
9514 1404 393
Answer:
L'(4, -3)
Step-by-step explanation:
The reflection over the horizontal line y=-1 effects the transformation ...
(x, y) ⇒ (x, -2-y)
So, the coordinates of L are transformed to ..
L(4, 1) ⇒ L'(4, -2-1) = L'(4, -3)
Answer:
C. 4.6
Step-by-step explanation:
If you subtract 3.0 from 1.4 you get 1.6, which is the can be added to 3.0 to find the missing value.
