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2 years ago

9

Ratna , who is in Mumbai , telephones her brother Ravi in Ahmedabad and speaks for 89 seconds . She is charged Rs 1.20 for every

16 seconds or part of 16 seconds . There is also a service charge of Rs 2 / - . Study the bill carefully . How much will it cost Ratna if she speaks to Ravi for 3 minutes ?
A) 14.40
B) 16.40
C) 18.40
D) 20.40

Please give explanation of your answer , please .

1 answer:

blondinia [14]2 years ago

7 0

Answer:

16.40

Step-by-step explanation:

3 min-> 180 s

180÷16≈11

11+1=12

12x1.20=14.40

14.40 + 2 = 16.40

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Area of a base is 2 3/4 and the height is 4/5 what is the volume

Serhud [2]

Answer:

2 4/20

Step-by-step explanation:

4 × 2 = 8

8 + 3 = 11

11/4 × 4/5

11 × 4 = 44

4×5= 20

44/20

Simplified is

2 4/20

8 0

2 years ago

Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 4 in the manner described. (Ent

QveST [7]

Answer:

$x=2\cos(t)$ and $y=-2\sin(t)+1$

Step-by-step explanation:

$(x-h)^2+(y-k)^2=r^2$ has parametric equations:

$(x-h)=r\cos(t) \text{ and } (y-k)=r\sin(t)$ .

Let's solve these for x and y respectively.

$x-h=r\cos(t)$ can be solved for x by adding h on both sides:

$x=r\cos(t)+h$ .

$y-k=r \sin(t)$ can be solve for y by adding k on both sides:

$y=r\sin(t)+k$ .

We can verify this works by plugging these back in for x and y respectively.

Let's do that:

$(r\cos(t)+h-h)^2+(r\sin(t)+k-k)^2$

$(r\cos(t))^2+(r\sin(t))^2$

$r^2\cos^2(t)+r^2\sin^2(t)$

$r^2(\cos^2(t)+\sin^2(t))$

$r^2(1)$ By a Pythagorean Identity.

$r^2$ which is what we had on the right hand side.

We have confirmed our parametric equations are correct.

Now here your h=0 while your k=1 and r=2.

So we are going to play with these parametric equations:

$x=2\cos(t)$ and $y=2\sin(t)+1$

We want to travel clockwise so we need to put -t and instead of t.

If we were going counterclockwise it would be just the t.

$x=2\cos(-t)$ and $y=2\sin(-t)+1$

Now cosine is even function while sine is an odd function so you could simplify this and say:

$x=2\cos(t)$ and $y=-2\sin(t)+1$ .

We want to find $\theta$ such that

$2\cos(t-\theta_1)=2 \text{ while } -2\sin(t-\theta_2)+1=1$ when t=0.

Let's start with the first equation:

$2\cos(t-\theta_1)=2$

Divide both sides by 2:

$\cos(t-\theta_1)=1$

We wanted to find $\theta_1$ for when $t=0$

$\cos(-\theta_1)=1$

Cosine is an even function:

$\cos(\theta_1)=1$

This happens when $\theta_1=2n\pi$ where n is an integer.

Let's do the second equation:

$-2\sin(t-\theta_2)+1=1$

Subtract 2 on both sides:

$-2\sin(t-\theta_2)=0$

Divide both sides by -2:

$\sin(t-\theta_2)=0$

Recall we are trying to find what $\theta_2$ is when t=0:

$\sin(0-\theta_2)=0$

$\sin(-\theta_2)=0$

Recall sine is an odd function:

$-\sin(\theta_2)=0$

Divide both sides by -1:

$\sin(\theta_2)=0$

$\theta_2=n\pi$

So this means we don't have to shift the cosine parametric equation at all because we can choose n=0 which means $\theta_1=2n\pi=2(0)\pi=0$ .

We also don't have to shift the sine parametric equation either since at n=0, we have $\theta_2=n\pi=0(\pi)=0$ .

So let's see what our equations look like now:

$x=2\cos(t)$ and $y=-2\sin(t)+1$

Let's verify these still work in our original equation:

$x^2+(y-1)^2$

$(2\cos(t))^2+(-2\sin(t))^2$

$2^2\cos^2(t)+(-2)^2\sin^2(t)$

$4\cos^2(t)+4\sin^2(t)$

$4(\cos^2(t)+\sin^2(t))$

$4(1)$

$4$

It still works.

Now let's see if we are being moving around the circle once around for values of t between $0$ and $2\pi$ .

This first table will be the first half of the rotation.

t 0 pi/4 pi/2 3pi/4 pi

x 2 sqrt(2) 0 -sqrt(2) -2

y 1 -sqrt(2)+1 -1 -sqrt(2)+1 1

Ok this is the fist half of the rotation. Are we moving clockwise from (2,1)?

If we are moving clockwise around a circle with radius 2 and center (0,1) starting at (2,1) our x's should be decreasing and our y's should be decreasing at the beginning we should see a 4th of a circle from the point (x,y)=(2,1) and the point (x,y)=(0,-1).

Now after that 4th, the x's will still decrease until we make half a rotation but the y's will increase as you can see from point (x,y)=(0,-1) to (x,y)=(-2,1). We have now made half a rotation around the circle whose center is (0,1) and radius is 2.

Let's look at the other half of the circle:

t pi 5pi/4 3pi/2 7pi/4 2pi

x -2 -sqrt(2) 0 sqrt(2) 2

y 1 sqrt(2)+1 3 sqrt(2)+1 1

So now for the talk half going clockwise we should see the x's increase since we are moving right for them. The y's increase after the half rotation but decrease after the 3/4th rotation.

We also stopped where we ended at the point (2,1).

3 0

2 years ago

Given the recursive formula find the first 4 terms for each. (not multiple choice) (MAFS.912.F-BF.1.3) A. An=an-1 3 when a1=5 B.

Neko [114]

Answer:

<h3>See explanations below</h3>

Step-by-step explanation:

1) Given the recursive function An=an-1 + 3 when a1 = 5, we are to find the first four terms;

First term a1 = 5

a2 = a1 +3

a2 = 5 + 3

a2 = 8

a3 = a2 + 3

a3 = 8+3

a3 = 11

a4 = a3 + 3

a4 = 11 + 3

a4 = 14

<em>The first four terms are 5, 8, 11 and 14</em>

<em></em>

<em>2) </em>For the recursive function An=an-1 + 2/3 when a1 = 1

a2 = a1 + 2/3

a2 = 1 + 2/3

a2 = 5/3

a3 = a2 + 2/3

a3 = 5/3 + 2/3

a3 = 7/3

a4 = a3 + 2/3

a4 = 7/3 + 2/3

a4 = 9/3

a4 = 3

<em>Hence the first four terms of the sequence are 2/3, 5/3, 7/3, 3</em>

<em></em>

3) For the recursive function An=an-1 + 12 when a1=30

a2 = a1 + 12

a2 = 30 + 12

a2 = 42

a3 = a2 +12

a3 = 42 + 12

a3 = 54

a4 = a3 + 12

a4 = 54+12

a4 = 66

<em>Hence the first four terms of the sequence are 30, 42, 54, 66</em>

3 0

2 years ago

WILL REWARD BRAINLIEST

Mashutka [201]

Answer:

Option A. is the correct option.

Step-by-step explanation:

If a number is divisible by 5 it is not necessary that the number will be divisible by 10.

For example: 25 is the number divisible by 5 but not divisible by 10.

Or in other words a number divisible by 5 should be even to be divisible by 10.

So for the conditional statement : A number is divisible by 10 if and only if it is divisible by 5 will be false because any one out of these two statements is false. We know biconditional is true only when both the statements are true or false means both the statements should have same truth value.

Therefore Option A. is the correct option.

8 0

2 years ago

Read 2 more answers

a scientist is studying the growth of bacteria the bacteria double every 20 min if there are 5 bacteria at the beginning how man

Pie

Answer:

15,625

Step-by-step explanation:

2h=120m

120/20 = 6

5^6 = 15,625

4 0

2 years ago