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vampirchik [111]
2 years ago
12

In Byron's class, there are 2 boys for every 3 girls. Today 7 new boys were added making the number of boys and girls the same.

How many boys and girls were there originally?​
Mathematics
1 answer:
jarptica [38.1K]2 years ago
8 0

Answer:

12

Step-by-step explanation:

2+3=5+7=12

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Jika f(x-1)=2x+5 dan g1(x-2)=4x+2 maka (g•f)1(0)
spayn [35]
G (x) = x + 1 (fog) (x) = x 2 + 3x + 1 ⇒ (fog) (x) = x 2 + 3x + 1 ⇒ f (g (x)) = x 2 + 3x + 1 ⇒ f (x + 1) = x 2 + 3x + 1 Eg x + 1 = p, then x = p - 1. ⇒ f (p) = (p - 1) 2 + 3 ( p - 1) + 1 ⇒ f (p) = p 2 - 2p + 1 + 3p - 3 + 1 ⇒ f (p) = p 2  + p - 1 So f (x) = x 2  + x - 1 - -> 
                                            ANSWER IS : x 2 + x - 1 
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3 years ago
Show the tens fact you used. Write the difference.
Mars2501 [29]
What exactly do you mean? Is there answer's to the problem or more to the problem?
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3 years ago
Help me answer this question
borishaifa [10]
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7 0
2 years ago
Help please and don't delete the question
Ahat [919]

Answer:

x = 2

Step-by-step explanation:

8(4-x) = 7x + 2

1) Distribute : 32 - 8x = 7x + 2

2) Subtract 32 : -8x = 7x - 30

3) Subtract 7x : -15x = -30

4) Divide 15 : x = 2 :)

4 0
2 years ago
Read 2 more answers
. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i
KIM [24]

Answer:

Error of Andrew: Made incorrect factors from the roots

Step-by-step explanation:

Roots of the polynomial are: 3, 2 + 2i, 2 - 2i. According to the factor theorem, if a is a root of the polynomial P(x), then (x - a) is  a factor of P(x). According to this definition:

(x - 3) , (x - (2 + 2i)) , (x - (2 - 2i)) are factors of the required polynomial.

Simplifying the brackets, we get:

(x - 3), (x - 2 - 2i), (x - 2 + 2i) are factors of the required polynomial.

This is the step where Andrew made the error. The factors will always be of the form (x - a) , not (x + a). Andrew wrote the complex factors in form of (x + a)  which resulted in the wrong answer.

So, the polynomial would be:

(x - 3)(x - 2 - 2i)(x - 2+2i)=0\\\\ (x-3)(x^{2}-2x+2xi-2x+4-4i-2xi+4i-4i^{2})=0\\\\ (x-3)(x^{2}-4x+4+4)=0\\\\ (x-3)(x^{2}-4x+8)=0\\\\ x^{3}-4x^{2}+8x-3x^{2}+12x-24=0\\\\ x^{3}-7x^{2}+20x-24=0

7 0
3 years ago
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