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3 years ago

What is the x and y intercepts Of “ -2y=3x-6 “ ?

Mathematics

1 answer:

vesna_86 [32]3 years ago

5 0

$-2y=3x-6\ \ \ \ |:(-2)\\\\y=-1.5x+3$

It's a slope-intercept form where a slope = -1.5 and y-intercept = 3.

x - intercept: y = 0

Therefore we have the equation:

-1.5x + 3 = 0 |-3

-1.5x = -3 |:(-1.5)

x = 2

Answer: x-intercept = 2, y-intercept = 3

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What is the determinant of the coefficient matrix of this system? <br> 4x-3y=-8 <br> 8x-3y=12

gladu [14]

The coefficient matrix is build with its rows representing each equation, and its columns representing each variable.

So, you may write the matrix as

$\left[\begin{array}{cc}\text{x-coefficient, 1st equation}&\text{y-coefficient, 1st equation}\\\text{x-coefficient, 2nd equation}&\text{y-coefficient, 2nd equation} \end{array}\right]$

which means

$\left[\begin{array}{cc}4&-3\\8&-3\end{array}\right]$

The determinant is computed subtracting diagonals:

$\left | \left[ \begin{array}{cc}a&b\\c&d\end{array}\right]\right | = ad-bc$

So, we have

$\left | \left[\begin{array}{cc}4&-3\\8&-3\end{array}\right] \right | = 4(-3) - 8(-3) = -4(-3) = 12$

8 0

3 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

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

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

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

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

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

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

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

Recall that:

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

Then

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

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

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

3 0

3 years ago

What does a2+b2=c2 mean ps yo you will get 20 points

Lynna [10]

That is for Geometry. So Pythagorean Theorem: a2 + b2 = c2. The Pythagorean Theorem is a formula that gives a relationship between the sides of a right triangle The Pythagorean Theorem only applies to RIGHT triangles. A RIGHT triangle is a triangle with a 90 degree angle.

6 0

3 years ago

Z1 and Z2 are vertical angles. Z2 has measure of 93° what is the measure of Z1? enter your answer in the box.

Soloha48 [4]

Answer:

m∠Z1 = 93°

Step-by-step explanation:

Vertical Angles are the angles opposite each other when two lines cross.

They are congruent (the same as each other).

Therefore, m∠Z1 = m∠Z2, so if m∠Z2 = 93° then m∠Z1 = 93°

6 0

2 years ago

Find the angle measure D FIRST ANSWER GETS BRAINIEST

asambeis [7]

Answer:

its 44

Step-by-step explanation:

180=46+90+x

180=136+x

44=x

5 0

3 years ago