CAN SOMEONE HELP ME PLEASE (ASAP)

Fifteen years later, a man will be two times as old as he was 15 years ago. How old is he now?

Inga [223]

Answer:

Let the present age of the man be x years.

Age of the man 15 years ago =(x−15) years and

Age of the man 15 years later =(x+15) years

According to the given condition, we have

(x+15)=2(x−15)

⇒x+15=2x−30

⇒x−2x=−30−15

⇒−x=−45 or x=45 years

Hence, the present age of the man is 45 years.

Step-by-step explanation:

Hope it helps you..

Your welcome in advance..

(◍•ᴗ•◍)

6 0

2 years ago

Read 2 more answers

How do you answer these two questions?

-BARSIC- [3]

I think you could multiply or divide. Hope this helps:)

7 0

3 years ago

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

Which of the following are solutions to the equation sinx cosx = 1/4? Check all that apply.

N76 [4]

The solution is B. π/12+nπ

proof
sinx cosx = 1/4 is equivalent to 2 sinx cosx = 1/2 or sin2x =1/2
so 2x = arcsin(1/2) = π/6 + 2nπ, so x = π/12+nπ

8 0

3 years ago

Read 2 more answers

Mean, Median, Mode and Range. Easy hard math. Please show work and give answers . Need help appreciate you all

tigry1 [53]

Answer:

:)

Step-by-step explanation:

quiz scores:

a. arrange them all from least to greatest to figure out the median first

your quiz: 15, 16, 17, 18, 19 middle # = 17 median = 17

friend's quiz: 12, 13, 17, 20, 20 middle # = 17 median = 17

now for mean add all of your quiz scores up and divide by how many #s

15 + 16 + 17 + 18 + 19 = 85 / 5 = 17 mean = 17

now your friend's:

12 + 13 + 17 + 20 + 20 = 82 / 5 = 16.4 mean = 16.4

mode is which number is used the most

your mode = you don't have a mode your friend's mode = 20

b. you have a higher mean than your friend

world populations:

a. range is to subtract the smallest number from the largest

803 - 31 = 772

b. mean is adding it all up and dividing by how many numbers there are:

803 + 487 + 348 + 368 + 730 + 31 = 2767 / 6 = 461.17

median is arranging them in order and finding the middle #

31, 348, 368, 487, 730, 803

368 + 487 = 855 / 2 = 427.5

mean = 461.17

median = 427.5

mode = no mode

tomato plants:

a. range is subtracting the smallest from the largest

52 - 36 = 16

b. mean is adding all the #s and dividing by how many #s there are:

36 + 45 + 52 + 40 + 38 +41 + 50 + 48 = 350 / 8 = 43.75

median is the middle from least to greatest

36, 38, 40, 41, 45, 48, 50, 52

41 + 45 = 86 / 2 = 43

mean = 43.75

median = 43

mode = no mode

hope this helps! :)

4 0

3 years ago