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klasskru [66]
3 years ago
13

1. Pia drew a circle with a circumference of C and a diameter of 14 in. Pia knows that , and she wrote the following equation to

represent the value of .
(a) There is an error in Pia’s equation. Write an equation to correctly represent the value of . Show your work.
(b) Write an equation and find the circumference of the circle.
Mathematics
2 answers:
Arada [10]3 years ago
6 0

♡ The Questions ♡

-Pia drew a circle with a circumference of C and a diameter of 14 in. Pia knows that C= πd, and she wrote the following equation to represent the value of π.

(a) There is an error in Pia’s equation. Write an equation to correctly represent the value of π.

(b) Write an equation and find the circumference of the circle.

Show your work.

-Hank used a semicircle, a rectangle, and a triangle to form the following composite figure. What is its area in square centimeters? Use 3.14 to approximate π. Show your work.

-At a zoo, the lion pen is surrounded by a ring-shaped sidewalk. The outer edge of the sidewalk is a circle with a radius of 11m. The inner edge of the sidewalk is a circle with a radius of 9m.

(a) Write and simplify an expression for the exact area of the sidewalk.

(b) Find the approximate area of the sidewalk. Use 3.14 to approximate π. Show your work.

* ୨୧ ┈┈┈┈┈┈┈┈┈┈┈┈ ୨୧*

♡ The Answer(s) ♡

-1 (A) --> N/A!

-1 (B) --> N/A!

-2 (A) --> 1,156 Sq Meters

-3 (A) --> 40π Sq Meters

-3 (B) --> 125.6 m^2

*୨୧ ┈┈┈┈┈┈┈┈┈┈┈┈ ୨୧*

♡ The Explanation(s)/Step-By-Step(s) ♡

-1 (A) --> N/A!

-1 (B) --> N/A!

-2 (A) --> We can figure out that the rectangle is 50 cm, and the triangle is 26 cm! In order to figure out the circle's cm, we can use the formula π x r^2. The radius of the circle is 10 cm, and the π is 3.14. We can multiply 3.14 by 10, this makes the number 31.4 cm! If we add the numbers, we end up with the sum of 107.4 cm. Converting 107.4 in Sq Meters leaves us with 1,156 Sq Meters! Therefore, 1,156 Sq Meters is your answer!

-3 (A) --> 125.6 m^2 is equal to 40π Sq Meters, This helps determine the exact area.

-3 (B) --> The approximate area of the sidewalk is 125.6 m^2. To solve this, you need to subtract the area of the inner edge of the lion pen from the outer edge of the lion pen.

A = π ( (11 m)^2 - (9 m)^2 )

A= π ( 121 m^2 - 81 m^2 )

A = π ( 40 m^2 )

A= 40π m^2. The area of the sidewalk comes out as 40π square meters, this is equal to 125.6 m^2, therefore, your answer will be 125.6 m^2.

*୨୧ ┈┈┈┈┈┈┈┈┈┈┈┈ ୨୧*

♡ Tips ♡

-Use some formulas!

-Use videos to help!

*୨୧ ┈┈┈┈┈┈┈┈┈┈┈┈ ୨୧*

♡ Formulas ♡

A = π x r^2 (Radius)

A = 1/2 B x H (Height and Base, Triangle)

A = B x H (Height and Base, Parallelogram)

A = S1 X S2 (Sides, Square)

erastovalidia [21]3 years ago
3 0

Answer:

did you figure out the answer?

Step-by-step explanation:

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Solve the simultaneous equation 2p - 3q = 4, 3p + 2q = 9. <br>b. if 223= 87 find x<br>​
wolverine [178]

Answer:

Step-by-step explanation:

Given the simultaneous equation 2p - 3q = 4 and 3p + 2q = 9, to get the value of p and q we will use elimination method.

2p - 3q = 4 ...................... 1 * 3

3p + 2q = 9 ..................... 2 * 2

Multiplying equation 1 by 3 and 3 by 2:

6p - 9q = 12

6p + 4q = 18

Subtracting both equation

-9q-4q = 12-18

-13q = -6

q = -6/-13

q = 6/13

Substituting q = 6/13 into equation 2

2p - 3(6/13) = 4

2p - 18/13 = 4

2p = 4+18/13

2p = (52+18)/13

2p = 70/13

p = 70/26

p = 35/13

<em>Hence p = 35/13 and q = 6/13</em>

<em></em>

<em>b) </em>If if 223ₓ = 87 find x

Using the number base system and converting 223ₓ  to base 2 will give us;

223ₓ = 2*x² + 2*x¹ + 3*x⁰

223ₓ  = 2x²+2x+3

​

Substituting back into the equation, 2x²+2x+3 = 87

2x²+2x+3-87 = 0

2x²+2x-84 = 0

x²+x-42 = 0

On factorizing:

(x²+6x)-(7x-42) = 0

x(x+6)-7(x+6) = 0

(x+6)(x-7) = 0

x+6 = 0 and x-7 = 0

x = -6 and 7

<em>Hence the value of x is 7 (neglecting the negative value)</em>

5 0
3 years ago
Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2
kifflom [539]

Answer:

The solution of the differential equation is y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

<u>i) Using characteristic equation:</u>

The characteristic equation method assumes that y(t)=e^{rt}, where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

y'=re^{rt}

y"=r^{2}e^{rt}

r^{2}e^{rt}+e^{rt}=0

(r^{2}+1)e^{rt}=0

As e^{rt} could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}

Using Euler's formula:

y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]

y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)

y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)

The particular solution of the differential equation is given by:

y(t)_{p}=ASin(2t)+BCos(2t)

y'(t)_{p}=2ACos(2t)-2BSin(2t)

y''(t)_{p}=-4ASin(2t)-4BCos(2t)

So we use these derivatives in the differential equation:

-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)

-3ASin(2t)-3BCos(2t)=Sin(2t)

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)

Using the initial conditions we can check that C1=5/3 and C2=2

<u>ii) Using Laplace Transform:</u>

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]  

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]=\frac{2}{(s^{2}+4)}

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) =\frac{2}{(s^{2}+4)}

(s²+1)·Y(s)-2s-1=\frac{2}{(s^{2}+4)}

(s²+1)·Y(s)=\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}

Y(s)=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}

Y(s)=\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}

Using partial franction method:

\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

Y(s)=\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}

Y(s)=-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[-\frac{1}{3} \frac{2}{s^{2}+4}]-ℒ⁻¹[2\frac{s }{s^{2}+1}]+ℒ⁻¹[\frac{5}{3}\frac{1}{s^{2}+1}]

y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)

3 0
3 years ago
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