All categories

2 years ago

У= 3х-5 у = 3х+12 Please help ASAP for my lil sister

Mathematics

1 answer:

exis [7]2 years ago

4 0

Answer:

Step-by-step explanation:

The equation is undefined.

You might be interested in

Evaluate 13-0.5w+6x when w=10 and x=1/2

Rus_ich [418]

Ok so first you need to put w and x in the equation —-> 13-0.5(10)+6(1/2) and now you just solve -0.5 times 10 is -5 so 13-(-5)+6(1/2) now multiply 6 times 1/2 which is 3 so now we have 13-(-5)+3= 21

6 0

3 years ago

Read 2 more answers

Find the equation of the line that passes through (1,3) and is perpendicular to y = 1 − 2 x

egoroff_w [7]

Answer:

$y=\displaystyle\frac{1}{2} x+\displaystyle \frac{5}{2}$

Step-by-step explanation:

Hi there!

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept.

Perpendicular lines always have slopes that are negative reciprocals (ex. 1/2 and -2, 3/4 and -4/3)

Determine the slope (m):

$y = 1 -2x$

Rearrange into slope-intercept form:

$y = -2x+1$

Now, we can identify clearly that the slope is -2. Because perpendicular lines always have slopes that are negative reciprocals, a perpendicular line would have a slope of $\displaystyle\frac{1}{2}$ . Plug this into $y=mx+b$ :

$y=\displaystyle\frac{1}{2} x+b$

Determine the y-intercept (b):

$y=\displaystyle\frac{1}{2} x+b$

Plug in the given point (1,3) and solve for b:

$3=\displaystyle\frac{1}{2} *1+b\\\\b=\displaystyle \frac{5}{2}$

Therefore, the y-intercept is $\displaystyle \frac{5}{2}$ . Plug this back into $y=\displaystyle\frac{1}{2} x+b$ :

$y=\displaystyle\frac{1}{2} x+\displaystyle \frac{5}{2}$

I hope this helps!

8 0

2 years ago

Find the length of abc express in terms of pie

Hitman42 [59]

You should choose c because it makes the most sense out of all of them

6 0

3 years ago

What eigen value for this matix (1 -2) (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$