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ANEK [815]
4 years ago
12

Help please

Mathematics
2 answers:
stiv31 [10]4 years ago
6 0
Adult ticket is $6.50 and student is $4.25. 6.50-4.25=2.25
6.50X 230=1495
4.25X180=765
1495+765=$2260
Leni [432]4 years ago
5 0

Answer:

8.065 is how much each adult payed

Step-by-step explanation:


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Maria invests $6,154 in a savings account with a fixed interest rate of 8% compounded continuously. What will the account balanc
Vitek1552 [10]

Answer:

$13,695.98

Step-by-step explanation:

We can use the continuous compound interest formula to solve:

A = Pe^{rt}

<em>P = principal amount</em>

<em>r = interest rate (decimal)</em>

<em>t = time (years)</em>

<em />

First, lets change 8% into a decimal:

8% -> \frac{8}{100} -> 0.08

Now, lets plug in the values:

A=6,154e^{0.08(10)}

A=13,695.98

The account balance after 10 years will be $13,695.98

3 0
3 years ago
Carol is standing 8 feet from a lamppost. The angle of elevation from her eyes to the top of the lamppost is 20o, and the angle
kupik [55]

Answer:

Height of the lamppost = 5.602 ft + 2.912 ft = 8.514 ft

Step-by-step explanation:

The illustration form a triangle that can be demarcated into 2 right angle triangle. One triangle representing depression triangle and the other elevation triangle.

Depression triangle

The opposite side of the triangle formed is the length of the pole from the base to the horizontal line of sight. Therefore,

using tangential ratio

tan 35° = opposite/adjacent

tan 35° = a/8

cross multiply

a = 8 tan 35°

a = 8 × 0.70020753821

a = 5.60166030568

a = 5.602 ft

Elevation triangle

The opposite side of this right angle triangle represent the length from the horizontal line of sight to the top of the lamppost.

tan 20° = opposite/adjacent

tan 20° = b/8

cross multiply

b = 8 tan 20°

b = 8 × 0.36397023426

b = 2.91176187413

b = 2.912 ft

Height of the lamppost = 5.602 ft + 2.912 ft = 8.514 ft

8 0
3 years ago
Which ordered pair is a solution to the system of inequalities?<br> y&gt;−2 and x+y≤4
Kamila [148]

Answer:-2,-3

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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
The side of a square is 3 and 5/2 inches
Murrr4er [49]
Perimeter=22 inches
Area=121/4 inches
7 0
3 years ago
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