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2 years ago

I need help please take a look at this

Mathematics

2 answers:

Reika [66]2 years ago

5 0

Answer:

$y = x-1$

$-x+\dfrac32=-y\\\implies y=x-\dfrac32$

The equations have the same slope but different y-intercepts. This means the lines are parallel and there is no solution (since they do not intersect).

Masteriza [31]2 years ago

5 0

$\\ \rm\hookrightarrow y=x-1$ --(1)

$\\ \rm\hookrightarrow -x+3/2=-y$

$\\ \rm\hookrightarrow y=x-3/2$ --(2)

Observe

Only b is different in both ones(y intercept)

Lines are parallel so no solution

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Which of the following rotational symmetry is applied to the regular not

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Answer:

Step-by-step explanation:

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3 years ago

Consider the isosceles triangle. left side (2x+8)units bottom of triangle (4x-10)units right side of triangle (2x+8) units Part

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Complete Question

Consider the isosceles triangle. left side (2z+8)units, bottom of triangle (4z-10)units, right side of triangle (2z+8) units Part A Which expression represents the perimeter of the triangle? a.(4z+16) units b.(6z−2)units c.(8z−16) units d.(8z+6) units

Answer:

d. (8z + 6) units

Step-by-step explanation:

The formula for the Perimeter of a Triangle is :Side A + Side B + Side C

Hence,

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Collect like terms

= 2z + 4z + 2z + 8 - 10 + 8

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3 years ago

Ms. Nickel wants to divide her class of 23 students into 4 equal teams . Is this reasonable ?

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3 years ago

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3 years ago

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What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

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3 years ago