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strojnjashka [21]
3 years ago
6

What is the correct factorization of x2 + 5x – 6?PICK THE RIGHT ONE (x – 1) (x + 6) (x + 1) (x – 6) (x – 2) (x – 3) (x – 2) (x +

3)
Mathematics
1 answer:
Arisa [49]3 years ago
7 0

Hey there!!

Given equation :

... x² + 5x - 6

... x² + 6x - x - 6

... x ( x + 6 ) - 1 ( x + 6 )

... ( x - 1 ) ( x + 6 ) is the answer.

Hope my answer helps!!

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The diagonals of rectangle nopq intersect at point r. if qr=3x-4 and np=5x+20, solve for x.
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Read 2 more answers
Can someone please help me ASAP.....
tester [92]

We have to start with a number x, compute f(x), and then use this result as input for g(x)

When you give x as input to f, it returns -x^2-1:

x\mapsto f(x)-x^2-1

Now, g(x)=x+5 simply means that g will return its input plus five.

In this case, the input is f(x), so we have

g(f(x))=f(x)+5 = -x^2-1+5=-x^2+4

5 0
4 years ago
There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl
Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

The number of students who can play only one of the clarinet, saxophone, or flute = 4.

6 0
4 years ago
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