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3 years ago

Which expression gives the distance between the points (-3,4 and (6,-2)

Mathematics

1 answer:

krok68 [10]3 years ago

8 0

Answer:

10.8

Step-by-step explanation:

We can find the distance using the distance formula:

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

We then substitute (-3,4) as $(x_1,y_1)$ and (6,-2) as $(x_2,y_2)$ .

$d=\sqrt{(6--3)^2+(-2-4)^2} \\d=\sqrt{(9)^2+(-6)^2} \\d=\sqrt{81+36}\\d=\sqrt{117}=10.8$

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ILL GIVE YOU BRAINLIEST (SMART MATH PERSON!)

MariettaO [177]

Answer:

A - (2a+0)/2 ; (2b+0)/2 = ( a , b )

B - (-2a+0)/2 ; (2b+0)/2 = ( - a, b )

C - (-2a+0)/2 ; ( -2b+0)/2 = ( - a, - b )

D - (2 a+0)/2 ; (-2b+0)/2 = ( a , - b )

Step-by-step explanation:

4 0

3 years ago

Find the solution of this system of equations<br> 4x - 5y = -12<br> -X + 5y = 18

Natalka [10]

4x - 5y = -12

-x + 5y = 18

Add the equations together:

4x + -x = 3x

-5y + 5y = 0

-12 + 18 = 6

So now you have:

3x = 6

Divide both sides by 3:

X = 2

Now you have the value of x, replace x with 2 in one of the equations and solve for y:

4(2) -5y = -12

8 -5y = -12

Subtract 8 from both sides:

-5y = -20

Divide both sides by -5:

Y = 4

The solution is x = 2 and y = 4 which is written as (2,4)

7 0

3 years ago

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p

GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

$t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2$

you replace t1 from the first equation in the second equation:

$(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}$

Then, for t2 = 0 and t2=1/2 you obtain for t1:

$t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2$

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths intersect are:

$r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)$

B) You have the following two trajectories of two independent particles:

$r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)$

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

$t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t$