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Colt1911 [192]
3 years ago
15

the slope of a line is –2 and its y-intercept is (0, 3). what is the equation of the line that is parallel to the first line and

passes through (2, 2)? a. y=4/3x - 3/2 b. 6x-4y=-8 c. y=3/4x 1 d. 4x – 2y = –12
Mathematics
1 answer:
Allisa [31]3 years ago
4 0
Parallel lines has equal slope.
Required equation is (y - 2)/(x - 2) = -2 => y - 2 = -2(x - 2) => y - 2 = -2x + 4 => 2x + y = 6
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Let g' be the group of real matricies of the form [1 x 0 1]. Is the map that sends x to this matrix an isomorphism?
aliina [53]

Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component x. So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.

However, let's prove in a more formal way that

\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]

is an isomorphism.

First of all, it is injective: suppose x \neq y. Then, you trivially have \phi(x) \neq \phi(y), because they are two different matrices:

\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]

Secondly, it is trivially surjective: the matrix

\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]

is clearly the image of the real number x.

Finally, \phi and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have

\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)

8 0
3 years ago
Lance bought 4 tickets to a concert each ticket cost $149 he sells tax rate for the tickets was 7.5% how much did Lance pay for
Vadim26 [7]
({149×4}÷100}×7.5 is the equation
3 0
3 years ago
Solve the system of equations.
KatRina [158]
Solve by Substitution
Answer:
x = -5
y = -2
5 0
2 years ago
the vertex of the parabola represented by f(x)=x^2-2x+6 has coordinates (1,5). Find the coordinates of the vertex of the parabol
Sever21 [200]
Plug \ in \ x+3 \ into \ f(x) \ to \ get \ g(x)
\\\ g(x) = (x+3)^2-2(x+3)+6
\\\ g(x)= x^2+6x+9 -2x-6+6
\\\ g(x) = x^2+4x+9
\\\ -\frac{b}{2a} --\ \textgreater \  vertex
\\\ \frac {-4}{2}
\\\ x \ coordinate = -2, plug \ it \ back \ into \ g(x)
\\\ (-2)^2+4(-2)+9
\\\ 4-8 + 9
\\\ y \ coordinate = 5
\\\ \boxed{(-2,5)}
5 0
3 years ago
Solve the system of linear equations below.<br><br> y = -2x + 19<br> y = x + 7
ASHA 777 [7]
Hey there!

y = -2x + 19
y = x + 7

We gonna solve this system of equation by using the substitution method.

We wanna solve y = -2x + 19 for y

Let start by substitute -2x + 19 for y in y = x + 7

y = x + 7
-2x + 19 = x + 7

Subtract x from both sides

-2x + 19 - x = x + 7 - x

-3x + 19 = 7

Now subtract 19 on both sides

-3x + 19 - 19 = 7 - 19

-3x = -12

Then divide both sides by -3

-3x/-3 = -12/-3

x = 4

We have the value of x. Now we gonna use that same value to find the value for y.

We gonna do that by substitute 4 for x in y = -2x + 19

y = -2x + 19

y = -2(4) + 19

y = -8 + 19

y = 11

Thus,

The answer is: x = 4 and y = 11

Let me know if you have questions about the answer. As always, it is my pleasure to help students like you!
6 0
3 years ago
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