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2 years ago

True or false: a closed circle is used when a number is a not solution

1 answer:

7 0

False

. A closed circle is used when a number is a solution. An arrow to the left or right shows that the graph continues in that direction.

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Functions h and k are inverse functions, and both are defined for all real numbers using this relationship, what is the value of

mestny [16]

Answer:

(h o k)(3) = 3

(k o h)(-4b) = -4b

Step-by-step explanation:

If f(x) and g(x) are inverses, then (f o g)(x) = (g o f)(x) = x

(h o k)(3) = 3

(k o h)(-4b) = -4b

5 0

2 years ago

Simplify the following expression

goldfiish [28.3K]

I believe the answer is : 33.94 , round if you need to

3 0

2 years ago

How do I find domain and range from a graph/equation?

likoan [24]

domain represents the x values so for example in a diagonal line that continues infinitely, the domain is all real numbers or (-infinity, infinity)

range represents y values so it would also be all real numbers or (-infinity, infinity)

let’s say there is a line (refer to pic) that moves ONLY from point (-3, -1) and (2, 2)

the domain would be [-3, 2]

we use brackets because it’s a real number unlike infinity (also because it’s a closed circle on the graph; if the graph had an open circle you would use a parenthesis)

and the range would be [-1, 2]

if you have any more questions about this explanation feel free to ask!

3 0

2 years ago

A train traveled 30 miles at an average speed of 60 mi/h ind the time taken for the trip

Ronch [10]

1/2 hour. 30/60 = 1/2 3 divided by 6 is 1/2 since it's not possible for it to be 1

4 0

3 years ago

Read 2 more answers

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$

we know that $cos^2A+sin^2A=1$ , so we have that

$f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)$

$lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19$

So f(0,0)=ln19.

8 0

3 years ago