The descriptions below outline different options for adding cards to your baseball card

Soni rent Anju a sum of Rs 40,000 at the rate of 2 paisa per 2 rupees per

galben [10]

Answer:

RS 14400

Step-by-step explanation:

Given the following :

Principal sum = RS 40000

Rate = 2 paisa per 2 rupee per month

Interest at the end of 3 years

1 rupee = 100 paisa

2 Paisa per 2 rupee per month :

2/200 per month = 1 / 100 per month,

Hence rate = 0.01 = 1% monthly

Annual rate = 1% × 12 = 12 % = 0.12

Simple interest :

Principal × rate × time

40000 × 0.12 × 3

40000 × 0.36

= RS 14,400

3 0

3 years ago

A bug completes one lap along the edge of a circular planter of radius 11.7 cm in 13.91 s. How fast was it traveling? 1. 10.8518

IrinaK [193]

Answer:

8.5.28492 cm/s

Step-by-step explanation:

We are given that

Radius of circular planter=11.7 cm

Time taken by bug =13.91 s

We have to find the speed of bug.

We know that

Speed = $\frac{distance}{time}$

Distance traveled by bug = $2\pi r$

Distance traveled by bug = $2\times 3.14\times 11.7$

Because $\pi=3.14$

Speed of bug= $\frac{2\times 3.14\times 11.7}{13.91}=5.28224 cm/s$

Hence, speed of bug=5.2822 4 cm/s

Answer:8.5.28492 cm/s

8 0

3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=3x%20%2B%20%28-2x%29%20%2B%202%20%3D" id="TexFormula1" title="3x + (-2x) + 2 =" alt="3x + (-2x

Oduvanchick [21]

The answer is 3x simplified
Evaluated 3x
Find the exact value is 3x

6 0

2 years ago

The area of a rectangle is 504 square meters. If the perimeter is 90 meters, find the length and width of the rectangle.

o-na [289]

Answer:

Length =24m and width = 21m

Step-by-step explanation:

Given,

Area of rectangle, A = 504 m^2

Perimeter of rectangle, P = 90 m

Since we know, A = Length * width and P = 2(Length + width)

So, A = L * w

=> 504 = L*w ------------------(1)

And P = 2(L + w)

=> 90 =2( L + w) ----------------(2)

Plug L =504/w in equation 2

i.e. 90 = 2( 504/w + w)

or 45 = 504/w + w

or 45 = (504 + w^2 ) / w

or 45w = 504 + w^2

or w^2 - 45w + 504 = 0

On solving above quadratic equation we get:

w = 21 or 24

So we take width , w= 21 meter

Then length will be, L =504/21 =24 meter

Hence length and width of rectangle are 24m and 21m respectively.

5 0

3 years ago