Please help, this is worth 15 points!
2 answers:
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Answer:
50
Step-by-step explanation:
since the Pythagorean theorem is
√
b is the bottom line (48)
a is the right line (14)
c is the top line (c)
plug that in
√
then do the exponets
= 14 * 14 = 196
= 48 * 48 = 2304
then add them together
196 + 2304 = 2500
√2500
the sqrt of 2500 is 50 (50 * 50 = 2500)
c = 50
your answer is 50
hope this helps:)
Answer:
(a)
(c)Maximum number of hours Lia can work at the restaurant and still meet her earnings goal = 15 hours
(d)Maximum Earning = $160
Step-by-step explanation:
Let r be the number of hours worked at the restaurant.
Let y be the number of hours of yard work,
Lia must work at least 5 hours per week in her family’s restaurant for $8 per hour.
Since she does yard work,
Lia’s parents allow her to work a maximum of 15 hours per week overall.
Lia’s goal is to earn at least $120 per week.
The restaurant, r pays $8 per hour
Yard work, y pays $12 per hour.
Therefore:
The system of inequalities that represent this problem is therefore:
(b)The graph of the inequality is attached below
(c)When the graph is plotted, the vertices of the feasible region are:
- (5,10)
- (5, 6.7)
- (15,0)
Where the first term is for the number of hours worked in the restaurant.
The maximum value of r possible is 15 from the three points.
Therefore, she can work at the restaurant for 15 hours and still meet her earning goal.
(d)Maximum Amount Lia can earn in 1 Week
- At (5,10), Earning=(5X8)+(10X12)=40+120=$160
- At (5, 6.7), Earning=(5X8)+(6.7X12)=40+80.4=$120.4
- At (15,0) Earning=(15X8)+(0X12)=$120
Since she has to work at least 5 hours at the restaurant, the maximum amount possible is $160.
Plug in m/4 into the equation:
Solve the parentheses in the equation:
Your equation should now look like this:
The answer is D.
Solve the parentheses in the equation:
Your equation should now look like this:
The answer is D.
Answer:
a. L{t} = 1/s² b. L{1} = 1/s
Step-by-step explanation:
Here is the complete question
The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0
Solution
a. L{t}
L{t} = ∫₀⁰⁰
Integrating by parts ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = and v =
and du/dt = dt/dt = 1
So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w
So, ∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + [
]₀⁰⁰
= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]
= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]
= -1/s[(0 - 0] - 1/s²[0 - 1]
= -1/s[(0] - 1/s²[- 1]
= 0 + 1/s²
= 1/s²
L{t} = 1/s²
b. L{1}
L{1} = ∫₀⁰⁰
= []₀⁰⁰
= -1/s[exp(-∞s) - exp(-0s)]
= -1/s[exp(-∞) - exp(-0)]
= -1/s[0 - 1]
= -1/s(-1)
= 1/s
L{1} = 1/s