All categories

3 years ago

2. Can a parallelogram have right angles?

Mathematics

2 answers:

LenKa [72]3 years ago

8 0

Answer:

yes

Step-by-step explanation:

because the parallelogram is equal to rectangle

motikmotik3 years ago

6 0

Answer:

yes

Step-by-step explanation:

A parolelogram is a rectangle and some rectangles have 4 right angles

You might be interested in

Wht is the volume of 729 cubed feet

lorasvet [3.4K]

Answer:

118085.765 hope this helps have a nice day please give me brainliest:)

Step-by-step explanation:

5 0

3 years ago

Sally is filling glass containers in the shape of cylinders with different colors of sand. Each cylinder has a radius of 4 inche

ryzh [129]

Part A.
We are given the total volume of the amount of sand she will want to fill
we are also given the shape and dimensions of the cylinders
Total volume of the sand = 1,000
Cylinder= radius 4 in, and height 8 in
the volume of a cylinder is V=πr^2h
We will solve for the volume of the cylinders
V=πr^2hV=π(4)^2*(10)
V = π (16) * (10)
V = π 160
The volume of the cylinders she wants to fill is 502.65 in^3
How many cylinders will she need
Well,
1000/502.65 ≈ 1.98
She will need two cylinder cans to fill with the 1,000 in^3 of sand

Part B.
To see if this is true, we find the half of the original cylinder's radius and height then solve for the volume and compare
V=πr^2h
The height and radius of the original cylinders were 4 r and 8h
we will find half of that which will be : 2 r and 4 h
Now solve for the volume V=πr^2h
V=π(2)^2* (4)
V=π (4)*(4)
V = π 16
V ≈ 50.26 inches ^3
The volume of the original cylinders was 502.65 in^3 and the volume of the new cylinders is 50.26 in^3... Clearly the volume of the new cylinders is NOT half of the original. Sally is not correct!

Hope this helps :)

7 0

3 years ago

Which of the following is a solution of x^2 + 2x + 4?

Kaylis [27]

Answer:

$\displaystyle x_1=-1+\sqrt{3}i$

$\displaystyle x_2=-1-\sqrt{3}i$

Step-by-step explanation:

Second-Degree Equation

The second-degree equation or quadratic equation has the general form

$ax^2+bx+c=0$

where a is non-zero.

There are many methods to solve the equation, one of the most-used is by using the solver formula:

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The equation of the question has the values: a=1, b=2, c=4, thus the values of x are

$\displaystyle x=\frac{-2\pm \sqrt{2^2-4\cdot 1\cdot 4}}{2\cdot 1}$

$\displaystyle x=\frac{-2\pm \sqrt{-12}}{2}$

Since the square root has a negative argument, both solutions for x are imaginary or complex. Simplifying the radical

$\displaystyle x=\frac{-2\pm 2\sqrt{-3}}{2}=-1\pm\sqrt{3}i$

The solutions are

$\displaystyle x_1=-1+\sqrt{3}i$

$\displaystyle x_2=-1-\sqrt{3}i$

7 0

3 years ago

Need help ASAP!!!!!!!

Temka [501]

Step-by-step explanation:

2/6 is litterally right there cause it has seven lines and tge 0 doesnt count

3 0

3 years ago

Find the area of the shaded portion of the figure. Each vertex of square ABCD is at the center of a circle. Round your answer to

IrinaK [193]

Given:

Each circle has a diameter of 2 inches each.

The outer square has a side length of 4 inches and the square ABCD has a side length of 2 inches.

To find:

The area of the shaded region.

Solution:

Each circle has a diameter of 2 inches. The square ABCD is at the center of each circle so it has a side length of 1 inch.

To determine the area of the shaded region, we subtract the area of the quarter-circles in the square ABCD from the area of the square ABCD.

The area of a quarter-circle $=\frac{\pi r^{2} }{4} .$

All the quarter-circles have a radius of 1 inch.

The area of 1 quarter-circle $=\frac{\pi r^{2} }{4} = \frac{\pi (1^{2}) }{4} = \frac{3.1415}{4} .$

The area of 4 quarter-circles $=4(\frac{3.1415}{4}) = 3.1415.$

So the area of the quarter-circles in the square ABCD is 3.1514 square inches.

The area of a square $= a^{2} .$

The area of square ABCD $=2^{2} =4.$

The area of the shaded region $=4-3.1415=0.8585.$

The area of the shaded region is 0.8585 square inches.

3 0

4 years ago

2. Can a parallelogram have right angles?​

2. Can a parallelogram have right angles?