All categories

2 years ago

Write the sum in standard form B) f(x)-h(x)

Mathematics

1 answer:

VLD [36.1K]2 years ago

8 0

Answer:

Step-by-step explanation:

f(x) -h(x) = x³ + 2x² -7x - 8 - [4x² - x - 15 ]

= x³ + 2x² - 7x - 8 - 4x² + x + 15

= x³ + 2x² - 4x² - 7x + x - 8 + 15

= x³ - 2x² - 6x + 7

You might be interested in

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

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

• Using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}$

4 0

3 years ago

Please help with this question

ELEN [110]

Answer:

r^2 * 1/s^4 * t^5

Step-by-step explanation:

r^2 * 1/s^4 * t^5

s^-4 = 1/s^4

7 0

3 years ago

How does a digit in the 10,000s place compare to a digit in the thousands place?

Delvig [45]

While a digit in the thousands place has a value 1,000 times the value of the digit. So to compare you can do 10,000 / 1,000 = 10, which means that a digit in the ten thousand place values ten times what the same digit values is it is the thousand place.

4 0

3 years ago

15t+30+5t=-4t+6 please show step by step

babunello [35]

15t+4t+5t=6-30
24t=-24
t=-1

3 0

3 years ago

Spiral Review (5.NBTA.1, 5.NBT A.2, 5.NBTB.5, 5.NET.B.6) 3. A small company packs 12 jars of jelly into each of 110 boxes to bri

marysya [2.9K]

If there are 12 jars of jelly in one box, and we have 110 boxes
we will do the following operation:
12 x 110=1320 jars of jelly

4 0

3 years ago