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4 years ago

1) 2+2+2=3----- 2) -----+------ = 3×6. 3) 8+-----+=------ ×8. use addition or multiplication

Mathematics

1 answer:

dolphi86 [110]4 years ago

8 0

Step-by-step explanation:

multiplication cause gives a reasonable answer.

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The sum of two polynomials is 10a2b2 – 8a2b 6ab2 – 4ab 2. if one addend is –5a2b2 12a2b – 5, what is the other addend? 15a2b2 –

Brrunno [24]

The other polynomial addend, when the sum of two polynomials is 10a2b2 – 8a2b 6ab2 – 4ab is 15a²b²- 20a²b + 6ab² - 4ab + 7.

<h3>What is polynomial?</h3>

Polynomial equations is the expression in which the highest power of the unknown variable is n (n is real number).

The sum of two polynomials is,

$10a^2b^2 - 8a^2b+ 6ab^2 - 4ab+2$

The one polynomial addend is,

$-5a^2b^2 +12a^2b - 5$

Let suppose the other polynomial addend is f(a,b). Thus,

$(-5a^2b^2 +12a^2b - 5)+f(a,b)=(10a^2b^2 - 8a^2b+ 6ab^2 - 4ab+2)$

Isolate the second polynomial as,

$f(a,b)=(10a^2b^2 - 8a^2b+ 6ab^2 - 4ab+2)-(-5a^2b^2 +12a^2b - 5)\\f(a,b)=10a^2b^2 - 8a^2b+ 6ab^2 - 4ab+2+5a^2b^2 -12a^2b + 5$

Arrange the like terms as,

$f(a,b)=10a^2b^2+5a^2b^2 - 8a^2b -12a^2b+ 6ab^2 - 4ab + 2+5\\f(a,b)=15a^2b^2- 20a^2b + 6ab^2 - 4ab + 7$

Hence, the other polynomial addend, when the sum of two polynomials is 10a2b2 – 8a2b 6ab2 – 4ab is 15a²b²- 20a²b + 6ab² - 4ab + 7.

Learn more about polynomial here;

brainly.com/question/24380382

7 0

2 years ago

40 50 60 70 80 90 100 110 120 130

RideAnS [48]

Answer:

um no problem with this is easy

5 0

3 years ago

Yoko runs 7 miles in 50 minutes at the same rate how many miles would she run in 75 minutes

Vsevolod [243]

Answer: 10.5 miles

Step-by-step explanation: If she runs 7 miles for 50 minutes, if you half it, she runs 3.5 miles in 25 minutes. This means if you add 7 and 3.5 miles, you would add 50 and 25 minutes,

So 10.5 miles in 75 minutes in your answer

6 0

3 years ago

Read 2 more answers

37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the

Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

(b) C is the boundary of the region

$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$