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nikdorinn [45]
2 years ago
14

Which one is a quadratic function

Mathematics
1 answer:
erastova [34]2 years ago
5 0

Answer:

Step-by-step explanation: A or the top answer choice. The highes exponent needs to be 2.

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36 – 0.0048 = ? A. 35.9952 B. 35.1520 C. 35.0048 D. 36.0048
Tems11 [23]
If you would like to solve 36 - 0.0048, you can calculate this using the following step:

36 - 0.0048 = 35.9952

The correct result would be A. 35.9952.
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What’s the Y and X Intercept <br> 5x+10y=10
vfiekz [6]

Answer:

X intercept =6

Y intercept =3

Step-by-step explanation:

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Crop yield is the ratio of the number of bushels harvested to the number of acres used for the harvest. This year, a large farm
Salsk061 [2.6K]

The amount of rice harvested this year from 690 acres of farmland was 5211570 bushels

<h3>What is an equation?</h3>

An equation is an expression used to show the relationship between two or more numbers and variables.

This year, a large farm harvested rice from 690 acres of farmland. The crop yield was 7,553 bushels per acre.

Hence:

Amount of rice harvested =  7,553 bushels per acre * 690 acres = 5211570 bushels

The amount of rice harvested this year from 690 acres of farmland was 5211570 bushels

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4 0
2 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
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