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3 years ago

Please help! I need these done before 2 pm!

Mathematics

1 answer:

jeyben [28]3 years ago

7 0

Answer:

9. (7a + 6b – 9c) – (3a – 6c)

=7a+6b-9c-3a+6c

=7a-3a+6b-9c+6c

=4a+6b-3c

10. (x2 – 9) – (-2x2 + 5x – 3)

= x^2-9+2x^2-5x+3

=x^2+2x^2-5x-9+3

=3x^2-5x-6

11. (5 – 6d – d2) – (-4d – d2)

=5-6d- d^2+4d+ d^2

=5-6d+4d-d^2+ d^2

=5-2d

12. (-4x + 7) – (3x – 7)

=-4x+7-3x+7

= -4x-3x+7+7

=-7x+14

13. (4a – 3b) – (5a – 2b)

=4a-3b-5a+2b

=4a-5a-3b+2b

= -a-b

14. (2c + 3d) – (-6d – 5c)

=2c+3d+6d+5c

=2c+5c+3d+6d

=7c+9d

15. (5x2 + 6x – 9) – (x2 – 3x +7)

=5x^2+6x-9- x^2+3x-7

=5x^2- x^2+6x+3x-9-7

=4x^2+9x-16

16. (3y – 6) – (8 – 9y)

=3y-6-8+9y

=3y+9y-6-8

=12y-14

17. (3a2 – 2ab + 3b2) - (-a2 – 5ab + 3b2)

=3a^2-2ab+3b^2+ a^2+5ab-3b^2

=3a^2+ a^2-2ab+5ab+3b^2- 3b^2

=4a^2+3ab

18. 5c – [8c – (6 – 3c)]

=5c-[8c-6+3c]

=5c-8c+6-3c

=5c-8c-3c+6

= -6c+6

19. 10x + [3x – (5x – 4)]

=10x+[ 3x-5x+4]

=10x+3x-5x+4

=8x+4

20. 3x 2 – [7x- (4x – x2) + 3]

=3x^2-[7x-4x+ x^2+3]

=3x^2-7x+4x- x^2-3

=3x^2-x^2-7x+4x-3

=2x^2-3x-3

21. x2 – [ - 3x+ ( 4 – 7x)]

= x^2-[ -3x+4-7x]

= x^2+3x-4+7x

= x^2+3x+7x-4

= x^2+10x-4

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Step-by-step explanation:

Given information:

Period = 24 hr

Maximum = 16 at t=16 hr.

Minimum = 10 at t=4 hr.

The general sin function is

$y=A+\sin(B(x-C))+D$ .... (1)

where, |A| is altitude, $\frac{2\pi}{B}$ is period, C is phase shift and D is midline.

Period is 24 hr.

$24=\dfrac{2\pi}{B}\Rightarrow B=\dfrac{\pi}{12}$

Altitude is

$A=\dfrac{Maximum-Minimum}{2}=\dfrac{16-10}{2}=3$

$D=\dfrac{Maximum+Minimum}{2}=\dfrac{16+10}{2}=13$

The function is minimum at t=4 and maximum at t=16,phase shift is

$C=\dfrac{16+4}{2}=10$

Substitute these values in equation (1).

$y=3\sin\left(\frac{\pi}{12}\left(x-10\right)\right)+13$

Therefore, the required function is $f(x)=3\sin\left(\frac{\pi}{12}\left(x-10\right)\right)+13$ .

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