Help please

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

2 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago

Answers:

ryzh [129]

Answers:

☆☆☆☆☆☆☆☆aTriangle ZYX
bTriangle YZX
cTriangle XZY
dTriangle XYZ

4 0

3 years ago

Y=1/3x+4 going through the point (9,-2)

Art [367]

Substitute x=9 and y=-2 into equation,
Left hand side=-2
Right hand side=1/3(9)+4= 3+4=7

Because LHS doesn’t = with RHS, the point (9,-2) doesn’t satisfy with the equation

Therefore the point doesn’t go through y=1/3x +4.

5 0

2 years ago

Can someone please answer this math problem for me, please answer all of them if you can

ohaa [14]

Answer:

Q5

∠3 = ∠4

Q6.

∠S + ∠T = 90°

Q7.

∠X + ∠Y = 180°

Q8.

∠1 = ∠4

Q9.

∠K = ∠L

7 0

3 years ago

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