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Elodia [21]
3 years ago
6

What is the area of this composite figure, in square centimeters? 176 209 242 264

Mathematics
2 answers:
otez555 [7]3 years ago
8 0

Answer:

THE ANSWERE IS B OR 209

Step-by-step explanation:

e-lub [12.9K]3 years ago
3 0

Answer:

209

Step-by-step explanation:

We can divide the figure into 2 parts to make this easier.

11x16=176 so the square is 176 square cm.

(11x6)/2 = 66/2 = 33 so the triangle is 33 square cm.

176+33 = 209 sq cm.

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If f(x) = 3x to the 2nd+ 1 and g(x) = 1-x, what is the value of (f-g)(2)?
melisa1 [442]

 

\displaystyle\\f(x)=3x^2+1\\g(x)=1-x\\\\(f-g)(x)=f(x)-g(x)=3x^2+1-(1-x)=3x^2+1-1+x=\boxed{3x^2+x}\\\\(f-g)(2)=f(2)-g(2)=3\cdot2^2+2=3\cdot4+2=12+2=\boxed{14}




3 0
3 years ago
Use the compound interest formula to compute the total amount accumulated and the interest eamed.
RideAnS [48]

Answer:

$12,137.39

Step-by-step explanation:

Use the Compound Amount formula:

A = P (1 + r/n)^(nt), where r is the interest rate as a decimal fraction, n is the number of times the interest is compounded each year, and t is the number of years.

Here, A = $9000(1 + 0.075/12)^(12*4), or

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7 0
3 years ago
Prove that sin3a-cos3a/sina+cosa=2sin2a-1
Sloan [31]

Answer:

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1

Step-by-step explanation:

we are given

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1

we can simplify left side and make it equal to right side

we can use trig identity

sin(3a)=3sin(a)-4sin^3(a)

cos(3a)=4cos^3(a)-3cos(a)

now, we can plug values

\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}

now, we can simplify

\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}

\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}

\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}

now, we can factor it

\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}

\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}

we can use trig identity

sin^2(a)+cos^2(a)=1

\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}

we can cancel terms

=3-4(1-sin(a)cos(a))

now, we can simplify it further

=3-4+4sin(a)cos(a))

=-1+4sin(a)cos(a))

=4sin(a)cos(a)-1

=2\times 2sin(a)cos(a)-1

now, we can use trig identity

2sin(a)cos(a)=sin(2a)

we can replace it

=2sin(2a)-1

so,

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1


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2 years ago
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Ronch [10]

Answer:

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Step-by-step explanation:

Factor  

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4 0
3 years ago
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