All categories

2 years ago

Write an equation of the line that passes through the given point with the given slope

Mathematics

1 answer:

Inessa05 [86]2 years ago

8 0

Did Yu Ever Find The Answer I got the same question??

You might be interested in

HELP FAST PLEASEEEEE

White raven [17]

Answer:

The third one

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Which is the solution to the equation solved for p? (q + p)q = (s + r)s

Brums [2.3K]

Answer:

Option B

Step-by-step explanation:

$(q + p)q = (s + r)s$

Lets open the brackets of L.H.S.

$=> q^2 + pq = (s+ r)s$

Lets substract q² from both the sides.

$=> q^2 + pq - q^2 = (s + r)s - q^2$

$=> pq = (s + r)s - q^2$

Lets divide both the sides by q.

$=> \frac{pq}{q} = \frac{(s + r)s - q^2}{q}$

$=> p = \frac{(s + r)s }{q} - \frac{q^2}{q} = \frac{(s + r)s}{q} - q$

7 0

2 years ago

Linear Equation <br> 24 + 6a = 6 - 4a <br> Please help me with this ASAP

KatRina [158]

Answer:

Step-by-step explanation:

1. Combine Like Terms

24 + 6a = 6 - 4a

+4a +4a

24 + 10a = 6

-24 -24

10a = -18

___ ___

10 10

a= -1.8

4 0

3 years ago

Find the system of inequalities which represents the shaded region on the coordinate plane.

dybincka [34]

Answer:

given, y≥2x+1 and

x−1

first, draw the graph for equations y=2x+1 and y=

x−1

for y=2x+1

substitute y=0 we get, 2x+1=0⟹x=−0.5

substitute x=0 we get, y=1

therefore, y=2x+1 line passes through (0.5,0) and (0,1) as shown in fig.

Hence, y≥2x+1 includes the region above the line.

for y=

x−1

substitute y=0 we get,

x−1=0⟹x=2

substitute x=0 we get, y=−1

therefore, y=2x+1 line passes through (2,0) and (0,-1) as shown in fig.

Hence, y>

x−1 includes the region above the line.

the intersection region is the shaded region as shown in above figure which includes I, II and III quadrants.

Therefore, quadrant IV has no solution please make as brainlelist

7 0

3 years ago

Consider the following planes. 3x − 2y + z = 1, 2x + y − 3z = 3

prisoha [69]

Solve for any of the variables; for instance, $z$ :

$z=1-3x+2y$

$z=-\dfrac{3-2x-y}3$

Then

$1-3x+2y=-\dfrac{3-2x-y}3\implies11x-5y=6\implies\begin{cases}y=\frac{11x-6}5\\z=1-3x+\frac{22x-12}5\end{cases}$

Let $x=t$ ; then the intersection is given by the vector-valued function

$\vec r(t)=\left(t,\dfrac{11t-6}5,1-3t+\dfrac{22t-12}5\right)$

$\vec r(t)=\left(t,\dfrac{11t-6}5,\dfrac{7t-7}5\right)$

3 0

3 years ago