What is the percent of increase from 3.7 to 5?

The table below shows two equations:

NikAS [45]

Answer:

I think it is the last one

Step-by-step explanation:

5 0

3 years ago

T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is

K(x)= $\frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}$

The given curve is y=7 $e^{x}$

${y}''=7e^{x}\\ {y}'=7e^{x}$

k(x)= $\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}$

${k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}$

For Maxima or Minima

${k(x)}'=0$

$7(e^x)(1+49e^{2x})(98e^{2x}-1)=0$

→ $e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0$

$e^{x}=0 , ∧ 1+49e^{2x}=0$ [not possible ∵there exists no value of x satisfying these equation]

→ $98e^{2x}-1=0$

Solving this we get

x= $-\frac{1}{2}\ln{98}$

As you will evaluate ${k(x})}''$ <0 at x= $-\frac{1}{2}\ln98$

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[ $-\frac{1}{2}\ln98$ ,1/√2]

k(x)= $\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}$

k(x)= $\frac{7}{\infty}$

k(x)=0

5 0

3 years ago

I

amid [387]

Answer:

21.56

Step-by-step explanation:

7 0

3 years ago

A toy rocket is launched vertically from 5 feet above ground level with an initial velocity of 112 feet per second. The height h

Masja [62]

Answer:

Step-by-step explanation:

a)The height h after t seconds is given by the equation h(t)=-16t^2+112t+5.

Where 5 represents 5feet above the ground and this is the height from which the rocket was launched.

The equation is a quadratic equation. The plot of this equation on a graph would give a parabola whose vertex would be equal to the maximum height travelled by the rocket.

The vertex of the parabola is calculated as follows,

Vertex = -b/2a

From the equation,

a = -16

b = 112

Vertex = - - 112/32= 3.5

So the rocket will attain maximum height at 3.5 seconds.

It will also take 3.5 seconds to reach the ground. This means it will take a total of 3.5 seconds + 3.5 seconds

= 7 seconds to hit the ground.

b) to find time it will take to be 100 feet above the ground,

-16t^2+112t-95 = 0

Using general quadratic equation formula,

a = -16

b = 112

c = -95

t = [-b+/-√b^2-4ac]/2a

t = [ -112 +/-√112^2-4×-16×-95]/16 × -2

= [-112 +/-√12544-6080]/-32

= (-112+80.4)/-32 or (-112-80.4)/-32

t = 0.9875 or t = 6.0125

So it will take 0.9875 or approximately 1 second to be 100 feet above the ground

7 0

3 years ago

Point A is located at (1, 5), and point M is located at (−1, 6). If point M is the midpoint of segment AB, find the location of

sashaice [31]

Ans: B=(-3, 7)

Step by step explanation:

7 0

3 years ago