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eimsori [14]
3 years ago
8

Find the area of each triangle?

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
3 0

Area of triangle A = 1/2bh
= 1/2 x 6 x 4
= 3 x 4
= 12 units

Area of triangle B = 1/2bh
= 1/2 x 4 x 3
= 2 x 3
= 6 units
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Expand the expression 3(7 + x)
Gre4nikov [31]

Answer:

3x + 21 or 21 + 3x

Step-by-step explanation:

Remark

This is an example of the distributive property. Of all the fundamental rules used in solving equations, this is likely the most important, at least at beginning levels.

Solution

What this is saying is two things.

1. you must multiply both terms on either side of the + sign by the factor outside the brackets.

2. You must not destroy a plus (or minus) sign unless the two terms inside the brackets are exactly the same.

3(7 + x)           The terms are not the same, so your answer must have a plus sign

Left side of the plus: 3*7

Right side of the plus: 3 * x

The answer then is 21 + 3x

It is more commonly written as 3x + 21

3 0
3 years ago
Find the area of a circle with diameter 18 ft
Semmy [17]
Hello : 
the area is : A=<span>πr²
</span> π ≃ 3.14     r : <span>ridus
</span>   <span>diameter = 2 r
</span>r = 18/2= 9 ft  
  A=3.14×9² = 254.34 ft²

8 0
3 years ago
Read 2 more answers
Michael is 3 times as old as Brandon. 18 years ago, Michael was 9 times as old as Brandon.
Zigmanuir [339]
I hope this helps you


Brandon x


Michael 3x


18 years ago


Brandon x-18


Michael 3x-18


9 (x-18)=3x-18


9x-9.18=3x-18

9x-3x=9.18-18

6x=8.18

x=8.3

x= 24
5 0
3 years ago
WILL GIVE A CROWN...A polygon is shown: A polygon MNOPQR is shown. The top vertex on the left is labeled M, and rest of the vert
strojnjashka [21]

The area of polygon MNOPQR = Area of a rectangle that is 15 square units + Area of a rectangle that is 2 square units.

In the given polygon MNOPQR, side MN is parallel to side RQ and the side MR is parallel to side PQ

We will draw a perpendicular line from point O on the side RQ, which will intersect RQ at point S. So, we can now divide the whole polygon into two different rectangles MNSR and OPQS with the areas as A₁ and A₂ respectively.

In rectangle MNSR, length(MN) = 5 units and width (MR) = 3 units

According to the formula for Area of rectangle,

A₁ = (length)×(width)

A₁ = (5 units)×(3 units)

A₁ = 15 square units

Now in rectangle MNSR, side MN= side RS and side MR = side NS,

so RS= 5 units and NS= 3 units

That means, SQ= RQ- RS = 7-5 = 2 units

and OS= NS - NO = 3- 2 = 1 unit

In rectangle OPQS, we have length(SQ) = 2 units and width(OS) = 1 unit

So, A₂ = (length)×(width)

A₂ = (2 units)×(1 unit)

A₂ = 2 square units

So, the area of polygon MNOPQR = (Area of a rectangle that is 15 square units + Area of a rectangle that is 2 square units)

4 0
4 years ago
Read 2 more answers
Brian's kite is flying above a field at the end of a 68m string.If the angle of elevation to the kite measures 70 degrees,how hi
Dvinal [7]

Answer:

h = 63.92\ m

Step-by-step explanation:

Given:

Angle of elevation = 70°

Length of string = 68 m

We need to find the height of the kite.

Solution:

Requirement figure attached in file.

Where:

BC = 68 m

AC = h (Height of the kite)

∠ABC = 70°

Using Cosine rule to find the height of the kite.

AC = BC\times sin(70)

h = 68\times 0.94

h = 63.92\ m

Therefore, height of the kite from Brian's head h = 63.92\ m.

4 0
4 years ago
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