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Leviafan [203]
3 years ago
5

Next week your math teacher is giving a chapter test. The test will consist of 35 questions. some problems r worth 2 point and s

ome r worth 4 points. there are 20 questions worth 2 points. how many problems are worth 4 points. solve by substitution
Mathematics
2 answers:
natima [27]3 years ago
8 0
There are 15 questions worth 4 points 
KATRIN_1 [288]3 years ago
5 0
There are 35 total questions. You already know that 20 of the questions are worth 2 points. You subtract 20 from 35. 35-20=15. So 15 questions are worth 4 points.
Total questions~35. Worth 2 points~20. Worth 4 points~15.
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Number line with open circle on 9 and shading to the left. Which of the following inequalities best represents the graph above?
yan [13]
Open or closed? because if it's closed then the the answer is D
3 0
3 years ago
There is a fraction that to 9/10. The numerator and the denominator is 57. What is the fraction?
xxTIMURxx [149]
The fraction is 27/30. It is equivalent to 9/10 and 27 + 30 = 57.
8 0
3 years ago
The area of a compact disc is 78.53 square centimeters. What is the radius of the disc? Use pi = 3.14
Tom [10]

Answer:

r = 5 cm

Step-by-step explanation:

A = πr²

78.53 = (3.14)r²

divide by 3.14

25 = r²

r = ±5

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5 0
3 years ago
The The Laplace Transform of a function , which is defined for all , is denoted by and is defined by the improper integral , as
guapka [62]

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} = ∫₀⁰⁰e^{-st}t

Integrating by parts  ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = e^{-st} and v = \frac{e^{-st}}{-s} and du/dt = dt/dt = 1

So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w

So,  ∫₀⁰⁰e^{-st}t =  [\frac{te^{-st}}{-s}]₀⁰⁰ -  ∫₀⁰⁰ \frac{e^{-st}}{-s}

∫₀⁰⁰e^{-st}t =  [\frac{te^{-st}}{-s}]₀⁰⁰ -  ∫₀⁰⁰ \frac{e^{-st}}{-s}

= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + \frac{1}{s} [\frac{e^{-st} }{-s}]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} = ∫₀⁰⁰e^{-st}1

= [\frac{e^{-st} }{-s}]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

6 0
3 years ago
00:00 Maggie is rock climbing. After reaching the summit, she descends 14 feet in 213 minutes. If she continues at this rate, wh
posledela
Set this up as ratio with feet on top and minutes on the bottom: \frac{14}{213}= \frac{x}{8}.  Cross multiply to get 213x = 112 and solve for x to get x = .52 feet.  This means that after she has been descending for 8 minutes she is a half a foot from the top.
6 0
3 years ago
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