All categories

2 years ago

N - 1/8 = 3/8 I am no smort and i forgot to study for a test =|

Mathematics

1 answer:

sdas [7]2 years ago

3 0

Answer:

$\boxed{\sf{n=\dfrac{1}{2}=0.5 }}$

Step-by-step explanation:

Isolate the term of n from one side of the equation.

First, add by 1/8 from both sides.

n-1/8+1/8=3/8+1/8

Solve.

Add the numbers from left to right.

3/8+1/8=4/8

Common factor of 4.

4/4=1

8/4=2

Rewrite as a fraction.

=1/2

n=1/2

Divide is another option.

1/2=0.5

n=1/2=0.5

Therefore, the final answer is n=1/2=0.5.

I hope this helps you! Let me know if my answer is wrong or not.

You might be interested in

What does 59% divide by 4 times 8 minus 9 times 5 equal?

Triss [41]

Answer: -43.82

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

I need help with this

miv72 [106K]

Answer:

y = -x + 3

Step-by-step explanation:

8 0

2 years ago

Which of the following is a point-slope equation of a line that passes through

Natasha_Volkova [10]

Answer:

y - 4 = -2/3 (x + 1)

Step-by-step explanation:

y2 - y1 / x2 - x1 -2 - 4 / 8 - (-1) -6/9 = -2/3

y - 4 = -2/3 (x + 1)

7 0

2 years ago

Consider f and c below. f(x, y) = (3 + 4xy2)i + 4x2yj, c is the arc of the hyperbola y = 1/x from (1, 1) to 3, 1 3 (a) find a fu

irga5000 [103]

We want to find a scalar function $f(x,y)$ such that $\nabla f(x,y)=\mathbf f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j$ .
So we need to have
$\dfrac{\partial f}{\partial x}=3+4xy^2$
Integrating both sides with respect to $x$ gives
$f(x,y)=3x+2x^2y^2+g(y)$
Differentiating with respect to $y$ gives
$\dfrac{\partial f}{\partial y}=4x^2y+\dfrac{\mathrm dg}{\mathrm dy}=4x^2y$ $\implies\dfrac{\mathrm dg}{\mathrm dy}=0$ $\implies g(y)=C$
So we find that
$f(x,y)=3x+2x^2y^2+C$
By the fundamental theorem of calculus, we then know the line integral depends only on the values of $f(x,y)$ at the endpoints of the path. Therefore

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f\left(3,\frac13\right)-f(1,1)=11-5=6$

4 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

2 years ago