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vladimir2022 [97]
3 years ago
12

A car traveled for 6 hours on one highway and for 2 hours on a second highway. If the total distance traveled was 488 miles, and

the car traveled 12 miles per hour faster on the second highway, at what average speed was the car traveling on the second highway?
Mathematics
1 answer:
Nataly [62]3 years ago
7 0

6V + 2(V+12) = 488
8V = 464
V = 58 mph for 6 hours

V+12=70 mph for 2 hours.


 check: 58*6 = 348

             70*2 = 140

348 +140 =488


 the car was traveling at 70 mph on the 2nd highway


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Perform the following computations.You may use13≈0.333333,34= 0.75 and100301= 0.332226.(i). Compute13+34by using five significan
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Answer:

a. 1.0833

Absolute Error = 0.416667

Relative Error = 1.250002

b. 0.0011070

Absolute Error = 0.0011070

Relative Error = 0.003321

Step-by-step explanation:

Given

1/3 = 0.333333

3/4 = 0.75

100/301 = 0.332226

a.

1/3 + 3/4

= 0.333333 + 0.75

= 1.083333

= 1.0833 ------ Approximated to 5 significant digits

Absolute Error = |Real Value - Estimated Value|

Relative Error = Absolute Error/Real Value

Assume 1/3 to be the real value and 3/4 to be the estimated value

Absolute Error = |0.333333 - 0.75|

Absolute Error = |-0.416667|

Absolute Error = 0.416667

Relative Error = 0.416667/0.333333

Relative Error = 1.250002

b.

1/3 - 100/301

= 0.333333 - 0.332226

= 0.001107

= 0.0011070 ----- Approximated to 5 significant digits

Assume 1/3 to be real value and 100/301 to be estimated value

Absolute Error = 0.333333 - 0.332226

Absolute Error = 0.0011070

Relative Error = 0.0011070/0.333333

Relative Error = 0.003321

Absolute and relative errors are approximation errors and they are due to the discrepancy between an exact value and some approximation to them.

The absolute error is the magnitude of the difference between the exact value and the approximation. The relative error is the absolute error divided by the magnitude of the exact value

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How do you know when a sequence is arithmetic vs. geometric?
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3 years ago
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Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

3 0
3 years ago
Expand using the properties and rules for logarithms
malfutka [58]

Consider expression \log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).

1. Use property

\log_a\dfrac{b}{c}=\log_ab-\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.

2. Use property

\log_abc=\log_ab+\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.

3. Use property

\log_ab^k=k\log_ab.

Then

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4. Use property

\log_{a^k}b=\dfrac{1}{k}\log_ab.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.

Answer: correct option is B.

7 0
3 years ago
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