Solve for t. −32=4t A.−8 B.−18 C.28 D.36

HELP ME SOMEONE IS CHASING ME HELP>>>>>

DENIUS [597]

Answer:

Suppose NOL -> QRM is a reflection across the x-axis. If O(-3, 8) what is the y coordinate of R?

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Is 5361 divisible by three how do you know just by looking at it

sleet_krkn [62]

Answer:

Yes!

Step-by-step explanation:

Here is a trick. If you add up all of the numbers in the number, and if that is divisible by three, the number is divisible by three. So...

5+3+6+1=15

15 is divisible by 3,

so 5361 is divisible by 3

(if you wanted to know, it is 1787)

8 0

2 years ago

Write the equation of the line with a slope of 3/7 passing through the point (9,-2) in standard or slope-point form? ...?

Musya8 [376]

$y-y_1=m(x-x_1)
\\
\\ m= \frac{3}{7} 
\\
\\(9,-2) \Rightarrow x_1=9,y_1=-2
\\
\\ y-(-2)=\frac{3}{7} (x-9)
\\
\\y+2=\frac{3}{7}x-\frac{27}{7}
\\
\\y=\frac{3}{7}x-\frac{27}{7}-2
\\
\\y=\frac{3}{7}x-\frac{41}{7}$

7 0

2 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

PLEASEEEEE SOMEBODY HELP NOW QUICK PLEASEEE

pickupchik [31]

Answer:

c

Step-by-step explanation:

becasue it would have a 3/8 chance of landing on silver , 8 goes into 240 30 times. so if we multiply 3 by 30 it gives us 90.

3 0

1 year ago