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Anvisha [2.4K]
3 years ago
9

The graph of linear function f passes through the point (3, 2) and has a slope of 2. What is the zero of f?

Mathematics
1 answer:
jeka943 years ago
4 0

Answer:

The graph of linear function f passes through the point (3, 2) and

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The boundary lines for the system of inequalities is given in the graph.
diamong [38]

Answer:

C)

region C

Step-by-step explanation:

We have to use what is called the zero-interval test [test point] in order to figure out which portion of the graph these inequalities share:

\displaystyle y ≤ -x + 2

0 ≤ 2 ☑ [We shade the portion of the graph that CONTAIN THE ORIGIN, which is the bottom portion.]

\displaystyle y ≥ 2x - 3

0 ≥ −3 ☑ [We shade the portion of the graph that CONTAINS THE ORIGIN, which is the left side.]

So, now that we got that all cleared up, we can tell that both graphs share a region where the ORIGIN IS VISIBLE. Therefore region C matches the above inequalities.

I am joyous to assist you anytime.

4 0
4 years ago
Read 2 more answers
Evaluate g(x)=3x when x= -2,0,and 5 g(-2)=? g(0)=? g(5)=?
serious [3.7K]

Answer:

g(-2) = -6

g(0) = 0

g(5) = 15

Step-by-step explanation:

For each of the evaluations, you have to plug in the number everywhere where there is an x

6 0
3 years ago
Who ever answers first gets branliest :))
scoundrel [369]
The answer is 36.86

(I think)
7 0
3 years ago
How do you solve break even problems?
Mandarinka [93]
A break even problem is found when you calculate starting a buisieness
so lets ssay you have to buy 3 coppy machines and each machine costs 2000 dollars,
the people who want to use your coppy machines have to pay $0.40 per page
so you have spent 6000 dollars already
when you break even, it is when your earnings equals your expenditure (how much you earned equals how much you paid)
8 0
3 years ago
Find limit as x approaches 2 from the left of the quotient of the absolute value of the quantity x minus 2, and the quantity x m
Molodets [167]

Answer:

\lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1.

Step-by-step explanation:

We want to find \lim_{x \to 2^-} \frac{|x-2|}{x-2}.

By definition:

|x-2|=\left \{ {{x-2,\:if\:x\:>\:2} \atop {-(x-2),\:if\:x\:

Since we want to find the Left Hand Limit, we use f(x)=-(x-2)

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} \frac{-(x-2)}{x-2}.

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} (-1).

The limit of a constant is the constant.

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1.

8 0
3 years ago
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