3. Find the domain and range of the function represented by the graph.
1 answer:
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Presumably we're after the line with that slope through that point. In general the line through 
with slope 
is
Here we have
Check: The slope is clearly
so we check
Here we have
Check: The slope is clearly
so we check
Answer: here it is graphed and it cant be solved because there are no solutions as seen on the graph!
Step-by-step explanation: hope i helped please mark brainliest!
Answer:
35
Step-by-step explanation:
The law of sines tells us ...
sin(C)/c = sin(A)/a
a·sin(3A) = c·sin(A)
Using the identity sin(3x) = 3cos(x)·sin(x) -sin(x)^3 and sin(x)^2 +cos(x)^2 = 1, we can simplify this to ...
sin(A)(4cos(A)^2 -1) = (c/a)sin(A)
4cos(A)^2 = c/a +1 = (48+27)/27 = 75/27 = 25/9
cos(A)^2 = 25/36
cos(A) = 5/6
__
Now, the angle B will be the difference between 180° and the sum of the other two angles:
B = 180° -A -3A = 180° -4A
Using appropriate trig identities, we can write ...
sin(B) = 4cos(A)^3sin(A) -4sin(A)^3cos(A)
= 4sin(A)cos(A)(cos(A)^2 -sin(A)^2)
= 4sin(A)cos(A)(2cos(A)^2 -1)
Filling in our value for cos(A), this becomes ...
sin(B) = 4sin(A)(5/6)(2(5/6)^2-1) = sin(A)(35/27)
__
The law of sines tells us ...
b/sin(B) = a/sin(A)
b = a·sin(B)/sin(A) = 27(35/27)sin(A)/sin(A) = 35
The length of side b is 35 units.
Answer:
For this case we can use the theorem of Existence and uniqueness that says:
Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:
has unique solution defined for all t in [a,b]
If we apply this to our equation we have that p(t) =0 and and
We see that is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by
And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.
Step-by-step explanation:
For this case we have the following differential equation given:
With the conditions y(1)= 1 and y'(1) = 7
The frist step on this case is divide both sides of the differential equation by t and we got:
For this case we can use the theorem of Existence and uniqueness that says:
Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:
has unique solution defined for all t in [a,b]
If we apply this to our equation we have that p(t) =0 and and
We see that is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by
And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.