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densk [106]
2 years ago
10

HELP0ppppppppppppppp​

Mathematics
1 answer:
Natalija [7]2 years ago
5 0

Answer:

C)

Step-by-step explanation:

in order to understand which triangle would lie on the given line we should figure out the slope of the given line first

remember that,

\displaystyle m =  \frac{ y_{2} -  y_{1} }{ x_{2} -  x_{1}}

from the graph let's take two points where the line passes i.e (0,0) and (4,6)

substitute:

\displaystyle m =  \frac{6-  0}{ 4 -  0}

simplify substraction:

\displaystyle m =  \frac{6 }{ 4 }

reduce fraction:

\displaystyle m =  \frac{3}{ 2}

now we need a triangle which tan ratio equal to the slope in that case we can consider the third triangle

\displaystyle \tan( \theta) =  \frac{30}{20}

reduce fraction:

\displaystyle \tan( \theta) =  \frac{3}{2}

since the tan ratio of the third triangle equal to the slope of the line

hence, our answer is c

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Help me please I need to do this
Blizzard [7]

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8 0
3 years ago
Evaluate f(4)<br> f(x) = 3x + 4
RSB [31]
The answer is definitely 7xf
7 0
2 years ago
I suck at math
solniwko [45]
Area of rectangle = length*breadth
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6 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
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scZoUnD [109]

Answer:

MNO no se estoy en 6th y no se pero esta es mi respuesta espero y me la saco bien

4 0
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