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3 years ago

What is the are of parallelogram ABDC?

Mathematics

2 answers:

SVEN [57.7K]3 years ago

6 0

Answer:

9 cm^2

Step-by-step explanation:

inside the parallelogram, two right angled triangles form.

Use the formula, A = $\frac{1}{2}$ x base x height (Area of a triangle) and multiply by 2 as there are two triangles. So,

A = ( $\frac{1}{2}$ x base x height) x 2

= ( $\frac{1}{2}$ x 3 x 3 ) x 2

= 9 cm^2

castortr0y [4]3 years ago

3 0

Answer:

9 cm2

Step-by-step explanation:

area=3×3=9cm2

.....

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A circle has a circumference of 7,805 units.

yulyashka [42]

Answer:

A circle has a circumference of 7,805 units.

2πr= 7,805 units.

r=7805/2π1=1242.20unit is radius

7 0

3 years ago

Factor 4x^4+8x^2+1<br> PLZ try hurry ;-;.....anyone?

Rudik [331]

The expression is not factorable with rational numbers.

4 0

3 years ago

The volume of a cylinder is 4 pi x3 cubic units and its height is x units.

Rzqust [24]

Answer:

$2x$

Step-by-step explanation:

The volume of a cylinder, $V$ is given by the formula,

$V=\pi r^{2}h$

Where, $r$ is the radius and $h$ is the height of the cylinder.

Here, $V=4\pi x^{3}$ , $h=x$ . Plug in these values and solve for radius, $r$ .

This gives,

$V=\pi r^{2}h\\4\pi x^{3}=\pi r^{2}x\\r^{2}=\frac{4\pi x^{3}}{\pi x}\\r^{2}=4x^{2}\\$

Taking square root both sides, we get

$\sqrt{r^{2}}=\sqrt{4x^{2}}\\r=2x$

Therefore, the radius of the cylinder can be expressed as $2x$ .

6 0

3 years ago

Describe what a graph of xs 1 would look like. (Be sure to include whether dot is open or closed and state

insens350 [35]

Answer:

The graph would be a line starting at one and moving to the left infinitely. The dot would be closed.

6 0

3 years ago

Please help with this and explain how to do it, I'll give 15 points

MrRissso [65]

Answer:

Option (4).

Step-by-step explanation:

Outer diameter of the hollow metallic ball = 10 centimeters

Outer radius of this ball = $\frac{10}{2}$ = 5 cm

Volume of the outer ball $V_{1}$ = $\frac{4}{3}\pi (r)^3$

= $\frac{4}{3}\pi (5)^3$

Inner radius of the hollow metallic ball = (5 - 1) = 4 cm

Volume of the inner hollow ball $V_{2}$ = $\frac{4}{3}\pi (4)^3$

Volume of the metal used in the metallic ball = $V_{1}-V_{2}$

= $\frac{4}{3}\pi (5)^3-\frac{4}{3}\pi (4)^{3}$

= $\frac{4}{3}\pi [(5)^3-(4)^3]$

Therefore, expression given in option (4) will be used to measure the volume of the hollow metallic ball.

8 0

3 years ago