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irina1246 [14]
3 years ago
6

Please help! 10 points

Mathematics
1 answer:
valina [46]3 years ago
4 0
3x^2+16x-35 is the answer and I need 20 word
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How do I do this I don’t get it
masya89 [10]

Answer: The answer would be 17/10.

Step-by-step explanation: You would divide Marvin's miles by Nathan's miles to find your answer.

8 0
3 years ago
If cos 0=2/3, what are the values of sin 0 and tan 0?
morpeh [17]

Answer:

Below

Step-by-step explanation:

● cos O = 2/3

We khow that:

● cos^2(O) + sin^2(O) =1

So : sin^2 (O)= 1-cos^2(O)

● sin^2(O) = 1 -(2/3)^2 = 1-4/9 = 9/9-4/9 = 5/9

● sin O = √(5)/3 or sin O = -√(5)/3

So we deduce that tan O will have two values since we don't khow the size of O.

■■■■■■■■■■■■■■■■■■■■■■■■■

●Tan (O) = sin(O)/cos(O)

● tan (O) = (√(5)/3)÷(2/3) or tan(O) = (-√(5)/3)÷(2/3)

● tan (O) = √(5)/2 or tan(O) = -√(5)/2

4 0
3 years ago
What is the explicit formula for this sequence?
djyliett [7]

Answer:

C. an= -3+ (n - 115)

Step-by-step explanation:

hope it help

5 0
3 years ago
The diagram shows a circle of radius 1 contained in a square. If the area of the circle equals x% of the area of the square, the
11111nata11111 [884]

Answer:

Area of square  with a side of 2r  = 2(1)  = 2  = 2^2   = 4

Area of circle =  pi ( 1)^2  =  pi

x =  pi / 4   ≈  .79   =  79%

7 0
3 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
3 years ago
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