What is the interest of $4,945 at 5% a month

Lily pads increase by 25% every 10 days.lf a pond starts with 22 lily pads,how many will there be after 50 days?

Fantom [35]

We conclude that after 50 days, there will be 67 lily pads.

<h3>how many will there be after 50 days?</h3>

We know that the number increases by 25% every 10 days.

In 50 days we have 5 groups of 10 days, so there will be five increases of 25%.

We know that the initial number is 22 lily pads, if we apply five consecutive increases of the 25% we get:

N = 22*(1 + 25%/100)*...*(1 + 25%/100%)

( the factor (1 + 25%/100%) appears five times)

So we can rewrite:

N = 22*(1 + 25%/100%)⁵

N = 22*(1 + 0.25)⁵ = 67.1

Which can be rounded to the nearest whole number, which is 67.

So we conclude that after 50 days, there will be 67 lily pads.

If you want to learn more about percentages:

brainly.com/question/843074

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3 0

2 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago

In this assignment, you will explore how the volume is changed when you adjust the height or radius of a cylinder. You will drag

lesantik [10]

The volume of a cylinder changes when you adjust the height or radius as the volume either increases or reduces.

<h3>How to illustrate the volume?</h3>

Let's assume that the height and radius are 14cm and 7cm. The volume will be:

= πr²h

= 3.14 × 7² × 14

= 2154.04cm³

When the radius and height are reduced to 5cm and 9cm, the volume will be:

= πr²h