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2 years ago

Can somebody solve this please

Mathematics

1 answer:

blondinia [14]2 years ago

4 0

What’s the question?

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2y-x=6 and y=2x+7 graphed equals how many solutions

zhannawk [14.2K]

Answer:

One solution

BRAINLIEST, PLEASE!

Step-by-step explanation:

2y - x = 6

2y = x + 6

y = x/2 + 3

y = 2x + 7

After graphing, they have one solution at (-2.667, 1.667).

8 0

2 years ago

The coordinates of the endpoints of AB and CD are A(2,

Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

$m= \frac{y_2-y_1}{x_2-x_1}$

Formula to Find Distance between two points

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

mAB ( represents , Slope of AB )

1. $mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}$

2. $mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}$

3. $mCD=\frac{5-2}{6-5}=\frac{3}{1}=3$

4. $AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

5. $BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is $-\frac{1}{3}$

$mAB=-\frac{1}{3}$

$mCD=3$

$mAB \times mCD = -\frac{1}{3} \times 3 = -1$

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0

3 years ago

Help pls asap -2x(-3x - 4y - 8) =

xenn [34]

-2x*(-3x)-2x*(-4y)-2x*(-8)

answer :6x^2+8xy+16x

3 0

2 years ago

a rectangular table top has a length of 4 3/4 ft and an area of 11 7/8 square feet what is the width of the table top

mina [271]

4¾=19/4 and 11⅞=95/8.
Width=area/length=95/8÷19/4=95/8×4/19=5/2=2½ft.

8 0

2 years ago

X = cos(2t) and y = sin(2t) Find the second derivative

Elodia [21]

Dy/dx = (dy/dt) / (dx/dt)
dy/dt = 2cos(2t)
dx/dt = -2sin(2t)

dy/dx = 2cos(2t)/-2sin(2t) = -cot(2t)

d^y/dx^2 = d(-1/tan(2t))/dx = (tan(2t) * 0 - (-1)*2sec^2(2t))/tan^2(2t) = 2sec^2(2t)/tan^2(2t) = 2csc^2 (2t)

3 0

3 years ago