Please help !!!!!!!!!!!

FIND THE SUM AND THE DIFFERENCE FOR EACH PAIR OF POLYNOMIALS please show work

Pepsi [2]

Answer:

12a. Addition = 7y⁵ + 7y³ – 6y – 5

12b. Subtraction = – y⁵ – 3y³ + 8y + 11

13a. Addition = 5x⁴ + x³ – 6x² + 8x + 17

13b. Subtraction = – x⁴ – 9x³ –6x² + 8x + 3

14a. Addition = – 2b⁴ + b³ – 11b² – 1

14b. Subtraction = 4b⁴ + 7b³ – b² + 8b – 1

15a. Addition = m⁵ + m⁴ + m³ + m²

15b. Subtraction = m⁵ – m⁴ – m³ – m² – 10

16a. Addition = 4x⁴ + 12x³ + 18x² + 16x + 4

16b. Subtraction = 4x⁴ – 4x³ + 14x² – 16x + 4

Step-by-step explanation:

12a. Addition

.. 3y⁵ + 2y³ + y + 3

+ (4y⁵ + 5y³– 7y – 8)

————————————

= 7y⁵ + 7y³ – 6y – 5

12b. Subtraction

.. 3y⁵ + 2y³ + y + 3

– (4y⁵ + 5y³– 7y – 8)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – y⁵ – 3y³ + 8y + 11

13a. Addition

.. 2x⁴ – 4x³ – 6x² + 8x + 10

+ (3x⁴ + 5x³ +...................+ 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 5x⁴ + x³ – 6x² + 8x + 17

13b. Subtraction

.. 2x⁴ – 4x³ – 6x² + 8x + 10

– (3x⁴ + 5x³ +...................+ 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – x⁴ – 9x³ – 6x² + 8x + 3

14a. Addition

...... b⁴ + 4b³ – 6b² + 4b – 1

+ (– 3b⁴ – 3b³ – 5b² – 4b)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – 2b⁴ + b³ – 11b² – 1

14b. Subtraction

...... b⁴ + 4b³ – 6b² + 4b – 1

– (– 3b⁴ – 3b³ – 5b² – 4b)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4b⁴ + 7b³ – b² + 8b – 1

15a. Addition

... m⁵ + m³ – 5

+ (m⁴ + m² + 5)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= m⁵ + m⁴ + m³ + m²

15b. Subtraction

... m⁵ + m³ – 5

– (m⁴ + m² + 5)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= m⁵ – m⁴ – m³ – m² – 10

16a. Addition

.... 4x⁴ + 8x³ + 16x² + 4

+ (4x³ + 2x² + 16x)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4x⁴ + 12x³ + 18x² + 16x + 4

16b. Subtraction

.... 4x⁴ + 8x³ + 16x² + 4

– (4x³ + 2x² + 16x)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4x⁴ – 4x³ + 14x² – 16x + 4

3 0

3 years ago

What is the Volume?

Serhud [2]

Answer:

703.36

Step-by-step explanation:

the formula is v=b×h×pi

pi = 3.14

length: 28

base:8

So base (8) times length (28) times pi (3.14)

8×28×3.14 = 703.36

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=0.07%20%5Ctimes%201.22" id="TexFormula1" title="0.07 \times 1.22" alt="0.07 \times 1.22" align

Alex_Xolod [135]

Answer:

0.0854

Step-by-step explanation:

$0.07 \times 1.22$ = 0.0854

Please mark brainliest. Have a great day!

3 0

3 years ago

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago

① 5J+2S =12.20 ② 5J+ 10S =15.80 elimination method

hram777 [196]

Answer:

Step-by-step explanation:

$5J +2S =12.20 ........(1)\\5J+10S=15.80......(2)\\Multiply \:equation \:(1)\:by\:the\:coefficient\:of\:x\:in equation\:(2)\\\\Multiply\:equation(2)\: by\: the\: coefficient\:of \:x\: in \:equation (1)\\\\5J +2S =12.20 ........(1) \times 5\\5J+10S=15.80......(2)\times 5\\\\25J+10S=61......(3)\\25J+50S = 79......(4)\\Subtract\:equation\:(4)\:from\: equation\: (3)\\-40S =-18\\\frac{-40S}{-40}=\frac{-18}{-40}\\ S = 9/20\\$

$Substitute \:9/20\:for \:x \:in\:equation\:(1)\:or \:equation\:(2)\\5J+2S = 12.20\\5J +2(9/20) =12.20\\5J +9/10=12.20\\5J=12.20-9/10\\5J=113/10\\Cross-Multiply\\50J = 113\\J = 113/50$

4 0

3 years ago