Find f(6) if f(x) = x2 ÷ 3 + x. 18 21 10 4

Which of the following have at least two congruent parallel bases? all that apply. A. Cylinder B. Circle C. Cone D. Cube E. None

Nookie1986 [14]

Answer:

Option A. Cylinder.

Option D. Cube

Step-by-step explanation:

In this question we have to tell that which are the figures have parallel bases.

Cylinder: Cylinder has two parallel congruent circular bases at the top and at the bottom.

Circle : It's a two dimensional structure and doesn't has any top or bottom.

Cone : It has only one circular bottom.

Pyramid : Pyramids have only one base circular or rectangular.

Cube : Cube is a three dimensional figure having minimum two congruent sides parallel to each other.

Therefore Option A. Cylinder and Option D. Cube are the answers.

7 0

2 years ago

Read 2 more answers

The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?

Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

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

Let,

$A(-6, 2)) = (x_1, y_1)$

$B(2, 6) = (x_2, y_2)$

$AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}$

$AB = \sqrt{(8)^2 + (4)^2}$

$AB = \sqrt{64 + 16} = \sqrt{80}$

$AB = 8.9$ (nearest tenth)

Distance between B(2, 6) and C(7, 1):

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

Let,

$B(2, 6) = (x_1, y_1)$

$C(7, 1) = (x_2, y_2)$

$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

$BC = \sqrt{(5)^2 + (-5)^2}$

$BC = \sqrt{25 + 25} = \sqrt{50}$

$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

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

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

$CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}$

$CD = \sqrt{(-4)^2 + (-6)^2}$

$CD = \sqrt{16 + 36} = \sqrt{52}$

$CD = 7.2$ (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

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

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

$DA = \sqrt{(-9)^2 + (7)^2}$

$DA = \sqrt{81 + 49} = \sqrt{130}$

$DA = 11.4$ (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0

2 years ago

Find the solution of the system of equations. 15x – 4y = -50 3x – 2y = –16

rusak2 [61]

Answer:

x=-2, y=5. (-2, 5).

Step-by-step explanation:

15x-4y=-50

3x-2y=-16

---------------

15x-4y=-50

-5(3x-2y)=-5(-16)

------------------------

15x-4y=-50

-15x+10y=80

-------------------

6y=30

y=30/6

y=5

3x-2(5)=-16

3x-10=-16

3x=-16+10

3x=-6

x=-6/3

x=-2

7 0

2 years ago

Math (2) Help please

eduard

What am i supposed to help with, can you show me?

8 0

3 years ago

find a parametric representation of the solution set of the linear equation. (enter your answer as a comma-separated list of equ

Ostrovityanka [42]

To represent the solution set of a linear equation parametrically, we introduce other parameters like s and t for the free variables.

Every linear equation has n - 1 free variables where n is the number of variables.

For x + y + z = 2, we have 3 variables and 3 - 1 = 2 free variables.

First, let y and z be the free variables, we first solve the linear equation for x to get:

x = 2 - y - z

Therefore , the parametric representation of the solution set is given by :

x = 2 - s -t

y = s

z = t

Learn more about The linear Equation at:

brainly.com/question/17748588

#SPJ4

5 0

1 year ago