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3 years ago

Which of the following expressions are polynomials?

Mathematics

1 answer:

AlekseyPX3 years ago

5 0

Answer:

1,3 and 5

Step-by-step explanation:

1) -2(7x + 1) − 5(3x2 + 3x + 2)

= -14x - 2 - 15x^2 - 15x - 10

= -15x^2 - 29x - 12

2) (5x2 − 10x + 8) − (10x2 + 19x + 20)

= 5x^2 - 10x + 8 - 10x^2 - 19x - 20

= -5x^2 -29x - 12

3) (-7x2 − 21x + 13) − (8x2 + 8x + 25)

= -7x^2 - 21x + 13 - 8x^2 - 8x - 25

= -15x^2 - 29x - 12

4) (-19x2 − 4x - 7) + (4x2 + 25x − 5)

= -15x^2 + 21x - 12

5) (-17x2 + 2x − 3) + (2x2 − 31x − 9)

= -15x^2 -29x - 12

6) -2(4x − 15) − 3(5x2 + 7x + 6)

= -8x + 30 - 15x^2 - 21x - 18

= -15x^2 - 29x + 12

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alina1380 [7]

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I rent a gym for $150 for 30 students another time I rent the gym for $270 for 70 students I need to find a fixed rate

kirza4 [7]

Answer:

Fixed rate is $60.

Step-by-step explanation:

Let us consider per student charge be 'x'.

Let us consider fixed rate be 'b'

Given:

I rent a gym for $150 for 30 students.

So we can say that;

Total amount is equal to sum of number of students multiplied by per student charge and fixed rate.

framing in equation form we get;

$30x+b=150 \ \ \ \ equation \ 1$

Also Given:

another time I rent the gym for $270 for 70 students.

So we can say that;

Total amount is equal to sum of number of students multiplied by per student charge and fixed rate.

framing in equation form we get;

$70x+b=270 \ \ \ \ equation \ 2$

Now we will subtract equation 1 from equation 2 we get;

$70x+b-(30x+b)=270-150\\\\70x+b-30x-b=120\\\\40x=120$

Dividing both side by 40 we get;

$\frac{40x}{40}=\frac{120}{40}\\\\x=\$3$

Now we will substitute the value of x in equation 1 we get;

$30x+b=150\\\\30\times3+b=150\\\\90+b=150\\\\b=150-90 =\$60$

So we can say that the equation can be written as;

$y =3x+60$

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Answer:

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