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3 years ago

What is the y-intercept of the following equation. 3x-2y=6

1 answer:

7 0

3x-2y=6

1) Add -3x to both sides.
 3x−2y=6
+3x +3x
-----------------
−2y=−3x+6

2) Divide both sides by -2.
 $\frac{-2y}{-2}$ = $\frac{−3x+6}{-2}$

y=32x−3

3) Substitute the y with the equation
3x-2(32x-3)=6
3x-64x+6=6

4) Combine like terms
3x-64x+6=6
-61x+6=6

5) Subtract 6 from each side of the equation
-61x+6=6
-6 -6 (+6-5 gets crossed out)
--------------
-61x=0

6) Divide both sides by -61
$\frac{-61x}{-61} = \frac{0}{-61}$
x=0

============================================================

Now that you've got you x intercept, use the x value and substitute it into the
y=32x−3

y=32(0)-3
y=0-3
y=-3

(0,-3)

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What is 58.65 with a 15% tip?

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X/58.65 = 15/100

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3 years ago

The slope of AB is -2/5. Segment CD has endpoints at C(-5,3) and D(10,-3). Are AB and CD parallel, perpendicular or neither? Jus

irina [24]

Answer:

Neither

Step-by-step explanation:

When I input segment CD onto a graph, I get -0.4 as the slope or 15/-6. If the lines were parallel, they would need to have the same slope which they do not have. If they were perpendicular, they would need to have different slopes so, -2/5 and 5/2 which they do not. That leaves us with the only option of neither.

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7 0

3 years ago

Read 2 more answers

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

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

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

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

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

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3 years ago

A. After Lou poured 2 liters of water into a large jug, the jug

Finger [1]

Answer:

8 liters

Step-by-step explanation:

If the sum of 2 + x = 10,

then you subtract 10 - 2 = x,

to get x = 8

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3 years ago

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Find the simple interest earned in a savings account that pays 5% interest if $500 is deposited for 4 years. A. $100, B. $1000,

g100num [7]

To calculate simple interest, you first need to multiply the original amount by the percentage of the money which is going to be added.

In this situation, 500 x 0.05 is the interest which will be added each year.
Therefore, $25 will be added each year. Since it is over 4 years, it is (25*4).
So the account will have earned $100 of simple interest.

Although the account will contain $600, it will have EARNED $100 of interest since it started with $500 already.

Answer - A $100 of simple interest will be earned.

3 0

3 years ago