Use the distributive property to remove the parentheses.

Passengers of flyaway airlines can purchase tickets for either business class or economy class. On one particular flight there w

Slav-nsk [51]

Answer:

x = 8

y = 146

Error = 4.86%

Step-by-step explanation:

Number of business class passenger = x

and the economy class passenger = y

Total number of passengers = 154

x + y = 154 ------(1)

Cost of business class tickets and economy class tickets are €320 and €85 respectively.

Total amount received by airlines is €14970.

320x + 85y = 14970

64x + 17y = 2994 --------(2)

Multiply equation (1) by 17 and subtract from equation (2)

(64x + 17y) - (17x + 17y) = 2994 - 2618

47x = 376

x = 8

From equation (1),

8 + y = 154

y = 154 - 8

y = 146

Airline officer wrote down the amount received as €14270

Then difference from the actual amount received = 14970 - 14270

= €700

% Error = $\frac{\text{Difference in amount}}{\text{Actual amount}}\times 100$

= $\frac{700}{14970}\times 100$

= 4.676

≈ 4.68%

Therefore, x = 8 and y = 146

and % error = 4.68%

8 0

2 years ago

Patty is using a cylinder shape container to hold 48 in³ of liquid. The height of the container is 6 inches tall. What is the be

Sati [7]

Answer:

Step-by-step explanation:

Volume of cylinder = πr2h

height =6 inches

r is the radius of the cylindrical container

48 in3 =πr2 * 6

6= πr2

π= 22/7

6*7/22=r2

42/22 =r2

21/11=r2

r2= 21/11

r = SQRT(21/11)

r= 1.38169855942

4 0

2 years ago

The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreas

iren [92.7K]

For the last part, you have to find where $f'(t)$ attains its maximum over $0\le t\le8$ . We have

$f'(t)=-6\sin3t+6\cos2t$

so that

$f''(t)=-18\cos3t-12\sin2t$

with critical points at $t$ such that

$-18\cos3t-12\sin2t=0$

$3\cos3t+2\sin2t=0$

$3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0$

$\cos t(3\cos^2t-9\sin^2t+4\sin t)=0$

$\cos t(12\sin^2t-4\sin t-3)=0$

So either

$\cos t=0\implies t=\dfrac{(2n+1)\pi}2$

or

$12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi$

where $n$ is any integer. We get 8 solutions over the given interval with $n=0,1,2$ from the first set of solutions, $n=0,1$ from the set of solutions where $\sin t=\dfrac{1+\sqrt{10}}6$ , and $n=1$ from the set of solutions where $\sin t=\dfrac{1-\sqrt{10}}6$ . They are approximately