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2 years ago

What is the answer to 4 1/10 -1 4/5

Mathematics

2 answers:

vagabundo [1.1K]2 years ago

8 0

2 and 3/10

if you need it as a decimal it would be 2.3

MrRissso [65]2 years ago

5 0

Answer:

2.3 is what I got :)

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Find the midpoint of the line segment whose endpoints are (-2, 5) and (6, -9). (4, -4) (2, -2) (1, -4)

Rus_ich [418]

Answer:

(2,-2)

Step-by-step explanation:

The midpoint of the line segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$ has coordinates

$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

If the endpoints of the line segment are (-2,5) and (6,-9), then the midpoint has the coordinates

$\left(\dfrac{-2+6}{2},\dfrac{5+(-9)}{2}\right)\\ \\\left(\dfrac{4}{2},\dfrac{-4}{2}\right)\\ \\(2,-2)$

8 0

3 years ago

Which of the following lines are parallel to 2Y - 3X = 4?

dimulka [17.4K]

Answer:

B. Y = 6/4 X

Step-by-step explanation:

Well to find its parallel line we need to put,

2y - 3x = 4 into slope-intercept.

+3x to both sides

2y = 3x + 4

Now we divide everything by 2,

y = 3/2x + 2

So a line that is parallel to the given line will have the same slope but different y intercept, meaning we can cross out choices A and C.

To check look at the image below ↓

Thus,

answer choice B. Y = 6/4 X is correct.

Hope this helps :)

4 0

3 years ago

Please help im dum.b

sleet_krkn [62]

Answer:

9/5

1 4/5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What are the coordinates of vertex C'

Harman [31]

Answer:

i what are u learning-

Step-by-step explanation:

8 0

2 years ago

Find a point on the ellipsoid x2+y2+4z2=36x2+y2+4z2=36 where the tangent plane is perpendicular to the line with parametric equa

Anvisha [2.4K]

Answer:

A point on the ellipsoid is (-4,2,2) or (4,-2,-2)

Step-by-step explanation:

Given equation of ellipsoid f(x,y,z) : $x^2+y^2+4z^2=36$

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

$\nabla f(x,y,z)=\\\nabla f(x,y,z)=$

So, The directions vectors=(-4,2,8)

Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

$\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)$

So, $2x=-4\lambda$

$\Rightarrow x=-2\lambda$

$2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda$

Substitute the value of x , y and z in the ellipsoid equation

$(2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2$

With $\lambda = 2$

x=-2(2)=-4

y=2

z=2

With $\lambda =- 2$

x=-2(-2)=4

y=-2

z=-2

Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)

7 0

3 years ago