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3 years ago

6

If the measure of arc XY plus the measure of arc YZX equals 360°, and the arcs do not overlap, then the arcs form a _____. A. po

lygon B. parallelogram C. square D. circle

2 answers:

s344n2d4d5 [400]3 years ago

5 0

I think it is a parallelogram

Brums [2.3K]3 years ago

5 0

Smh the answer is a circle.

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Prove that sin3a-cos3a/sina+cosa=2sin2a-1

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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

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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

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

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we can use trig identity

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

we can cancel terms

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

now, we can simplify it further

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

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

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

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

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

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2 years ago

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Simplify each expression then factor the expression 14-28t-5(7+10t)

ludmilkaskok [199]

Answer:

Simplified: -21-98t

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Step-by-step explanation:

The given expresion is:

14-28t-5(7+10t)

We expand the parenthesis to get:

$14 - 28t -35 - 70t$

We group like terms to get:

$14 -35 - 28t- 70t$

We combine similar terms to get:

$- 21- 98t$

We now factor to obtain:

$- 21- 98t = - 7(14t + 3)$

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