Given the function. y = 3x + 5
For, x = -3; y = 3(-3) + 5 = -9 + 5 = -4
For x = 1; y = 3(1) + 5 = 3 + 5 = 8
For x = 4; y = 3(4) + 5 = 12 + 5 = 17
Thus the table representing the function is the table with: -3, 1 and 4 as x-values and -4, 8, 17 as y-values.
For x = 0; y = 3(0) + 5 = 0 + 5 = 5
For y = 0; 0 = 3x + 5; 3x = -5 and x = -5/3
Thus the graph of the function is a straight line passing through points (0, 5) and (-5/3, 0).
To illustrate the fuction as a word statement we say that y is five more than three times x
From the given descriptions, the graph does not represent the graph of y = 3x + 5.
Therefore, the one that does not describe the same situation is the graph.
4 cos² x - 3 = 0
4 cos² x = 3
cos² x = 3/4
cos x = ±(√3)/2
Fixing the squared cosine doesn't discriminate among quadrants. There's one in every quadrant
cos x = ± cos(π/6)
Let's do plus first. In general, cos x = cos a has solutions x = ±a + 2πk integer k
cos x = cos(π/6)
x = ±π/6 + 2πk
Minus next.
cos x = -cos(π/6)
cos x = cos(π - π/6)
cos x = cos(5π/6)
x = ±5π/6 + 2πk
We'll write all our solutions as
x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k
I think But I'm not 100% sure yet it's a guess
Y=3+x
Root: (-3, 0)
Intercept:(0, 3)
Answer:
Yes, F is a continuous function of r
Step-by-step explanation:
We are given that
When r<R
Where M=Mass of the earth
R=Radius of earth
G=Gravitational constant
We have to find the function is continuous of r or not.
LHL
RHL
When a function is continuous at x=a
Then, LHL=RHL=f(a)
RHL==LHL=F(R)
Hence, the function is continuous of r.
