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3 years ago

Solve for the indicated variable in the literal equation

Mathematics

1 answer:

ANTONII [103]3 years ago

8 0

Answer:

$x = \frac{y}{3} - \frac{5}{3}$

Step-by-step explanation:

Given

$y = 5 + 3x$

Required

Solve for x

Subtract 5 from both sides

$y-5 = 5 -5+ 3x$

$y-5 = 3x$

Divide through by 3

$\frac{y}{3} - \frac{5}{3} = x$

$x = \frac{y}{3} - \frac{5}{3}$

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IrinaVladis [17]

All you have to do is divide each one by its other number. For example, 5.60 for 8 is the same as one for 0.70. We accomplish this by dividing the 5.60 by 8.
So...
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The first choice is the lowest.

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3 years ago

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snow_tiger [21]

Answer:

Step-by-step explanation:

7^-2 = 1/7^2

The negative means you flip it

1/7^2 times 7^6 is 7^6/7^2 (just multiply).

using rules of exponents, 7^6/7^2 is 7^4 (just subtract 2 from 6)

D is the only one that equals 7^4, as 2 times 2 is 4.

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3 years ago

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WITCHER [35]

Answer:

Step-by-step explanation:

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2 years ago

The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

lesya692 [45]

(1) D

$L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt$

Integrate by parts, taking

$u = t \implies \mathrm du=\mathrm dt$

$\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}$

Then

$L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}$

(2) A

$L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}$

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3 years ago

Solve for the indicated variable in the literal equation ​

Solve for the indicated variable in the literal equation