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3 years ago

What is the area of this figure?

Mathematics

2 answers:

finlep [7]3 years ago

8 0

Find the area of the 4 triangles.

12*10/2=60. There are two of the same dimensions, so the area so far is 120

12*18/2=108. There are also 2 of these, to the area of the bottom half is 216.
120+216=336.

lord [1]3 years ago

6 0

Two triangle areas of:

10*24/2+18*24/2

120+216

336 u^2

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ss7ja [257]

Answer:

R = 14 cm

A = pi R^2 = 3.14 " 196 = 616 cm^2

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2 years ago

Find the area of this polygon

Levart [38]

<h3>Answer: 12 square units</h3>

Explanation:

Rectangle ABDE is 4 units across the horizontal, and 2 units tall.

The area of this rectangle is length*width = 4*2 = 8 square units.

Triangle BCD has a base of 4 and height 2. The area of which is base*height/2 = 4*2/2 = 4 square units.

The total area is

rectangle + triangle = 8 + 4 = 12 square units

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2 years ago

0.0236 correct to 2 significant figure is

Oliga [24]

My key can’t perform this but I can solve it on a sheet...
So here I really hopes this helps

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2 years ago

Find the value of x. Round your answer to the nearest tenth

Sati [7]

$x=13.8$

<u><em>Step-by-step explanation:</em></u>

Since the triangle is right use the tangent ratio to solve for x

$tan41^o=\frac{opposite}{adjacent}=\frac{12}{x}$

Multiply both sides by $x$

x × $tan41^o=12$ ( divide both sides by $tan41^o$ )

x = ≈ ( nearest tenth )

6 0

2 years ago

A pitcher holds 3/4 gallons of kool-aid. Martin wants to pour equal amounts of kool- aid into 6 smaller cotainers. How much kool

statuscvo [17]

Answer:

$\dfrac{1}{8}\ gallons$

Step-by-step explanation:

Amount of kool - aid that the pitcher can hold = $\frac{3}4$ gallons

There are 6 smaller containers in which this kool - aid is to be adjusted.

Also, the volume of the smaller containers is equal to each other i.e. the containers are identical to each other in terms of volume.

To find:

The amount of kool - aid that each container can hold?

Solution:

Volume of all the smaller containers combined = Volume of the kool - aid available

Let the volume of one smaller container = $V$

$V + V + V + V + V + V = \dfrac{3}{4}\ gallons$

$\Rightarrow 6V = \dfrac{3}{4}\ gallons\\\Rightarrow V = \dfrac{3}{4\times 6 }\ gallons\\\Rightarrow V=\bold{\dfrac{1}{8}\ gallons}$

$\frac{1}{8}\ gallons$ of kool - aid is the amount that each smaller can hold.

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3 years ago