Need help on my work I'm marking you the brainiest if the answer is right! hurry pleaseeeeeeee need help quick asap! helpppppppp
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2 answers:
Remark
You have to start with the equality equations first and then go to the inequalities.
Step One
Discuss y = 2x + 2. That boundary is a dotted line. That is because it is not included in the area defined by y < 2x + 2. You should also note that everything in this region is below that line. See the diagram on the left (below).
Step Two
Discuss x + y >= -3. This graph actually means y >=-x - 3 The line itself is the lower boundary of the inequality. It looks like the center graph.
Step Three
When you put the two graphs together, you get the diagram on the right. The mauve kind of color is the region you are interested in. (Mauve is a kind of red purple).
Answer
Lower Right
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Answer:
(a) k'(0) = f'(0)g(0) + f(0)g'(0)
(b) m'(5) =
Step-by-step explanation:
(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use <u>product rule</u>,i.e.,
this will give <em>k'(x)=f'(x)g(x) + f(x)g'(x)</em>
on substituting the value x=0, we will get the value of k'(0)
{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}
(b) m(x) is a function of two functions f(x) and g(x) [ ]. Since m(x) has a function g(x) in the denominator so we need to use <u>division rule</u> to differentiate m(x). Division rule is as follows :
this will give <em></em>
on substituting the value x=5, we will get the value of m'(5).
{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}
{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }
{ NOTE : }
Answer: The correct is answer is: The subset consists of all the sandwiches with either white bread and ham or rye bread and turkey.
Step-by-step explanation: I got this question correct.
