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2 years ago

Help me out, thank you.

Mathematics

1 answer:

AveGali [126]2 years ago

7 0

Answer: 38

Step-by-step explanation:

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A demand ___ is a graphic representation of the law of demand

yan [13]

Answer:

I think it is b. Curve

Step-by-step explanation:

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3 years ago

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Lester’s car can go 15.4 miles on 1 gallon of gas. How far can he go on 0.7 gallon?

MAVERICK [17]

Answer:

He can go 10.78 miles

Step-by-step explanation:

This problem can be solve by creating a proportion, knowing that 15.4 miles is to one gallon the same as "x" miles (our unknown) is to 0.7 gallons. This can be set in math terms as the equation that relates these two ratios of miles per gallon:

$\frac{15.4\,miles}{1\,gallon} =\frac{x }{0.7\,gallon}$

and to solve for the unknown number of mile, we multiply both sides of the equation by "0.7 gallon" in order to isolate our unknown:

$\frac{15.4\,miles}{1\,gallon} =\frac{x}{0.7\,gallon}\\\frac{15.4\,miles\,*\,0.7\,gallon}{1\,gallon} =x \\x=10.78\,miles$

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3 years ago

Help I’m giving brainlest to the first person who get it right!!!

tensa zangetsu [6.8K]

Answer:

14,400 this is correct

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2 years ago

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Write a summary can Smartphones improve student lives? why or why not ( in your own words )

bonufazy [111]

Answer:

Smartphones can imrove student lives by giving students the ability to contact their teachers to ask questions or comment on a specific work. they can also allow the student to look up answers and learn from what others think. Smartphones can also add distractions in class and allow an easy way to cheat on a test or a quiz. Overall Smartphones can be beneficial and harmful to a students life.

Step-by-step explanation:

4 0

2 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

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

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