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FinnZ [79.3K]
3 years ago
10

What is the area of the composite figure?

Mathematics
1 answer:
ivolga24 [154]3 years ago
7 0

Answer:

745.4 cm^2

Step-by-step explanation:

The composite figure is made up of a rectangle 25 cm by 20 cm and a semicircle of radius 12.5 cm.

We add the two areas to find the total area.

total area = area of rectangle + area of semicircle

TA = LW + (1/2)(pi)r^2

TA = (25 cm)(20 cm) + (1/2)(3.14159)(12.5 cm)^2

TA = 500 cm^2 + 245.4 cm^2

TA = 745.4 cm^2

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Alchen [17]
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3 years ago
Please ASAP help please !!! Help me anyone
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difference of squares means that both the terms are square terms. (also there must be a - symbol)

for example

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so you would factorise it like this:

(y+2)(y-2)

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<em>h</em><em>o</em><em>w</em><em>e</em><em>v</em><em>e</em><em>r</em><em>,</em><em> </em>-2 doesnt have a whole number square root so it is not a difference of squares.

2. 25 has a square root of 5. m^2 has a square root of m. n^4 has a square root of n^2. so this 25m^2n^4 is a square term.

1 has a square root of +1 and -1.

therefore, this one is a difference of squares. <u>(</u><u>5</u><u>m</u><u>n</u><u>^</u><u>2</u><u> </u><u>+</u><u>1</u><u>)</u><u> </u><u>(</u><u>5</u><u>mn^2</u><u> </u><u>-</u><u>1</u><u>)</u>

3. p^8 has a square root of p^4. q^4 has a square root of +q^2 and -q^2)

so it is a difference of squares. <u>(</u><u>p</u><u>^</u><u>4</u><u>+</u><u>q</u><u>^</u><u>2</u><u>)</u><u>(</u><u>p</u><u>^</u><u>4</u><u> </u><u>-</u><u>q</u><u>^</u><u>2</u><u>)</u>

4. 16x^2 is a square term as irs square root is 4x.

<em>h</em><em>o</em><em>w</em><em>e</em><em>v</em><em>e</em><em>r</em><em>,</em><em> </em>24 is not a square term.

therefore, it is not a difference of squares.

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--------------------------  =  -------------------

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factor the numerator


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--------------------------  =  -------------------

cos^2 x                        1 - sinx


cos^2 = 1-sin^2x

(sinx +3) (sinx+1)       3 + sinx

--------------------------  =  -------------------

1- sin^2x                       1 - sinx

factor the denominator

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--------------------------  =  -------------------

(1-sinx ) (1+sinx)                   1 - sinx

cancel the common term (1+sinx)  and (sinx +1)

(sinx +3)                       3 + sinx

--------------------------  =  -------------------

(1-sinx )                           1 - sinx


reorder the first term

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