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4 years ago

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Mathematics

1 answer:

dedylja [7]4 years ago

5 0

Answer:

so...what is the rest of your question?

Step-by-step explanation:

soybeans-500 acres

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Which equation represents the line that passes through the point (4, -5) and is perpendicular to the line x + 2y = 5?

Lerok [7]

Answer:

y = 2x - 13

Step-by-step explanation:

Equation of a line is y = mx + c, m is the gradient and c is the intercept

The line passes through points 4 and -5, x is 4 and y is -5

-5 = 4m + c

When two lines are perpendicular, the products of their gradients are equal to -1, m1 * m2 = -1

x + 2y = 5

2y = -x + 5

y = (-1/2 * x) + 5

therefore m = -1/2

m1 * m2 = -1

m * -1/2 = -1

-m = -2 , therefore m = 2

-5 = 4 * 2 + c

c = -5 - 8, which is -13

Therefore the equation for the line is

y = 2x - 13

7 0

3 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago

Help me pleaseeee.........

ruslelena [56]

9514 1404 393

Answer:

y = 20
4^8
x = 0

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

a^0 = 1 . . . . for a ≠ 0

And, it is convenient to know the cubes of small integers:

1³ = 1; 2³ = 8; 3³ = 27; 4³ = 64; 5³ = 125

6³ = 216; 7³ = 343; 8³ = 512; 9³ = 729; 10³ = 1000

1) p^3 × p^5 = p^-12 × p^y

Equating exponents:

3 + 5 = -12 + y

20 = y . . . . . . . . . . . add 12

2) 64 × 4^5 = 4^3 × 4^5 = 4^(3+5) = 4^8

3) 10^x = 1 = 10^0

x = 0

6 0

3 years ago

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PLS PLS HELP ASAP<br> for a test

Nastasia [14]

Answer:

Airplane A is the slowest

Step-by-step explanation:

Airplane A = 400 m/s

Airplane B = 760 m/s

Airplane C = 740 m/s

Hope this helps!

6 0

3 years ago

How do i factor 2y^2+y-10

Leokris [45]

2y^2+y-10
(2y+5)(y-2)

4 0

3 years ago

Read 2 more answers