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Arisa [49]
3 years ago
13

Choose correct graph with following function

Mathematics
1 answer:
Vlad1618 [11]3 years ago
4 0

Answer:

The answer to your question is The second graph

Step-by-step explanation:

The general equation for a trigonometric is

                        Acos(Bx + C) + D

A indicates the amplitude if the value is 1, the lowest point is -1 and the highest point is 1.  So we discard the first graph.

B indicates the period. A normal graph has a period of 2π, but this equation has a period of 3/2. This means that the period is smaller.

C indicates that the graph is moved to the left or right but in this problem C = 0, so it does not change.

D indicates if the graph is moved up or down, in this problem, D = 0 so it does not change.

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The length breadth and height of a sectangular box are 24 cm, 6 cm and 2cm respectevely. Find the area of largest surface of box
slamgirl [31]

Answer:

408cmx^{2}

Step-by-step explanation:

A=2(wl+hl+hw)=2·(2·6+24·6+24·2)= 408cm²

6 0
2 years ago
The density of a fish tank is 0.4fish over feet cubed. There are 12 fish in the tank. What is the volume of the tank?
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Answer: 30\ ft.^3

Step-by-step explanation:

Given: Density of tank = 0.4 fish over feet cubed.

We know that, \text{Density}=\frac{\text{Mass}}{\text{Volume}}

Therefore,

\text{Volume of the tank}=\frac{\text{Mass}}{\text{Density}}\\\\\Rightarrow\ \text{Volume of the tank}=\frac{12}{0.4}\\\\\Rightarrow\ \text{Volume of the tank}=30\ ft.^3

Hence, the volume of the tank = 30\ ft.^3

4 0
3 years ago
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A cone has a diameter of 6 inches and a height of 6 inches. Find the volume of the cone to the nearest tenth. Use 3.14 for pi.
Oksi-84 [34.3K]

Answer:

The volume is 56.6 rounded

Step-by-step explanation:

6 0
3 years ago
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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
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4 years ago
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