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Feliz [49]
2 years ago
14

Use the information below to find RV.​

Mathematics
1 answer:
Inessa [10]2 years ago
5 0

Answer:

I think Rv = 37x but if you think not then Go with You're answer

Step-by-step explanation:

goodluck! :)

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Is 11/13 a ratio in simplest form
Molodets [167]
Yes 11/13 cannot be simplified anymore
5 0
3 years ago
Rite the first ten terms of a sequence whose first term is -10 and whose common difference is -2.
Lelechka [254]
-10, -10-2·1= -12, -10-2·2=-14, -10-2·3=-16, -10-2·4= -18, -10-2·5= -20,
 -10-2·6= -22, -10-2·7 =-24, -10-2·8=-26, -10-2·9=-28

a=-10
d=-2
xn=a+d(n-1)

the first ten terms are
-10, -12, -14, -16, -18, -20, -22, -24, -26, -28

8 0
3 years ago
Who ever gets this right will get a brainlest
zepelin [54]

Answer:

<1 and <2

<3 and <2

<1 and <4

Step-by-step explanation:

Adjacent angles are angles that are next to each other. In essence, they would share one aside in common. In this problem, the answers are the following,

<1 and <2

<3 and <2

<1 and <4

5 0
3 years ago
Read 2 more answers
What is the product of the polynomials below? (4x^2-2x-4)(2x+4)
Travka [436]

Answer:  8x³ + 12x² - 16x - 16

<u>Step-by-step explanation:</u>

  (4x² - 2x - 4)(2x + 4)

= (2x + 4)(4x² - 2x - 4)

= 2x(4x² - 2x - 4)  +  4(4x² - 2x - 4)

=   8x³ - 4x² - 8x   +  16x² - 8x - 16

=   8x³ + (-4x² + 16x²) + (-8x - 8x) - 16

=   8x³ + 12x² - 16x - 16

7 0
2 years ago
4 Tan A/1-Tan^4=Tan2A + Sin2A​
Eva8 [605]

tan(2<em>A</em>) + sin(2<em>A</em>) = sin(2<em>A</em>)/cos(2<em>A</em>) + sin(2<em>A</em>)

• rewrite tan = sin/cos

… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

… = 2/(2 cos²(<em>A</em>) - 1) (sin(<em>A</em>) cos(<em>A</em>) + sin(<em>A</em>) cos(<em>A</em>) (cos²(<em>A</em>) - sin²(<em>A</em>)))

• factor out sin(<em>A</em>) cos(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (1 + cos²(<em>A</em>) - sin²(<em>A</em>))

• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (2 cos²(<em>A</em>))

• rearrange terms in the product

… = 2 sin(<em>A</em>) cos(<em>A</em>) (2 cos²(<em>A</em>))/(2 cos²(<em>A</em>) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - 1/cos²(<em>A</em>))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

… = 4 tan(<em>A</em>) / (sec²(<em>A</em>) (2 - sec²(<em>A</em>)))

• rewrite using the Pythagorean identity, sec²(<em>A</em>) = 1 + tan²(<em>A</em>)

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (2 - (1 + tan²(<em>A</em>))))

• simplify

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (1 - tan²(<em>A</em>)))

• condense the denominator as the difference of squares

… = 4 tan(<em>A</em>) / (1 - tan⁴(<em>A</em>))

(Note that some of these steps are optional or can be done simultaneously)

7 0
2 years ago
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