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DanielleElmas [232]
4 years ago
12

How do I do this I don’t know how

Mathematics
1 answer:
larisa86 [58]4 years ago
7 0
In order to solve this problem, I would approach it in three steps:

1. First, calculate the area of the triangles drawn inside the polygon. It looks like each of the triangles has a base length of 6cm and a height of 8cm. The formula for the area of a triangle is 1/2×b×h.

Using this formula, we can calculate the area for one triangle to be:

1/2 × 6 × 8 = 24cm²

We also have a total of 8 triangles that each have the same base and height lengths, so we can find the total area that the triangles occupy by multiplying the area (24cm²) by 8.

So, the total area occupied by the triangles is:

8 × 24 = 192cm²


2. Second, we can calculate the area of the square inside the polygon. The formula for the area of a square is b*h. Our square has a base length of 12cm and a height of 12cm.

Using our formula, we can calculate the area of the square as:

12 × 12 = 144cm²


3. Our last step is to add our total areas together. This will give us the area of the whole shape.

192cm² + 144cm² = 336cm²



<em>Our final area for the polygon is 336cm².</em>




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Answer:

yes

Step-by-step explanation:

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3 years ago
Use the substitution method to solve the following system of equations: 3x + 5y = 3 x + 2y = 0 (1 point) (1, 1) (6, −3) (6, 1) (
PtichkaEL [24]

Step-by-step explanation:

Given.

3x+5y=3

x+2y=0

Take the second equation and subtract 2y to both sides.

(x+2y)-2y=0-2y

x=-2y

Substitute x into the first equation and simplify.

3x+5y=3

3(-2y)+5y=3

-6y+5y=3

-y=3

Invert.

y=-3

Substitute Y into your second equation.

x+2y=0

x+2(-3)=0

x-6=0

Add -6 to both sides.

(x-6)+6=0+6

x=6

Answer:

(6, -3)

6 0
3 years ago
Solve the following integral.<br><br> <img src="https://tex.z-dn.net/?f=%5Cint4x%5Ccos%282-3x%29dx" id="TexFormula1" title="\int
bagirrra123 [75]

Hi there!

\boxed{-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C}

To find the indefinite integral, we must integrate by parts.

Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":

u = 4x

du = 4

dv = cos(2 - 3x)

v = 1/3sin(2 - 3x)

Write into the format:

∫udv = uv - ∫vdu

Thus, utilize the solved for expressions above:

4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx

Simplify:

-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx

Integrate the integral:

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u = 2 - 3x

du = -3dx ⇒ -1/3du = dx

-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C

Combine:

-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C

7 0
2 years ago
Read 2 more answers
What is the simplified version for these 4 ?
VladimirAG [237]
1. -3x + -6
2. -3x + 9
3. 2x - 6
4. -2x + 6

here u go
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3 years ago
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10 is the missing number.
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