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Thepotemich [5.8K]
3 years ago
12

Metoda comparației

Mathematics
1 answer:
kozerog [31]3 years ago
5 0

Answer:

Aflam cat costa 1 stilou (38 - 8) : 6 = 30 : 6 = 5 lei

:)

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Which of the following points are solutions to the system of inequalities shown below?
vlada-n [284]
The following solutions to the given system of inequalities shown above would be option c : (-3, -5) option d : (0, 4)  option e. (4, 4)  and option f. <span>(2, -1) 
So how did I know the answer. To check this, you just have to substitute the values.
For example let us take option c (-3, -5)
y </span><span><  3x + 5
-5 </span><span><  3(-3) + 5
-5 </span><span><  -9 + 5
-5 </span><span><  -4 <<You see that -4 is greater than -5 which makes this inequality correct. This is the same process as with the other correct options.
Hope that this answer helps.</span>
7 0
3 years ago
Your parents gave you a gift card for your favorite coffee shop. The card was worth $50 when you got it. Each day, you buy a dri
Sever21 [200]

50=3x+b. you never specify the variable that replaces b

7 0
3 years ago
Which expression is a prime polynomial?
irakobra [83]

The expression that is a prime polynomial is:

B. x^3 - 27y^2.

<h3>What is a prime polynomial?</h3>

A prime polynomial is a polynomial that cannot be factored.

In this problem, item b gives a prime polynomial, as:

  • In item a, 3 is a common factor, hence the polynomial can be factored.
  • In item c, x is a common factor, hence the polynomial can be factored.
  • In item d, the polynomial can be factored according to it's roots.

More can be learned about prime polynomials at brainly.com/question/26388060

#SPJ1

4 0
2 years ago
The sum of 3 consecutive integers is 60. What is the value of the third integer?
Scorpion4ik [409]

Answer:

x=19

Step-by-step explanation:

The consecutive integers will be x, x+1 and x+2 since consecutive integers have a pattern.

So,  the equation is x+x+1+x+2=60

Solve:

x+x+1+x+2=60

3x+3=60

Subtract 3 on both sides:

3x+3-3=60-3

3x=57

Divide by 3

3x/3=57/3

x=19

3 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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