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4 years ago

I am terribly at geometry. Was always good at basic algebra. But not understanding this question

2 answers:

5 0

3x+y=15
First convert the equation into slope intercept form. y=mx+b
Where m is the slope and b is the y-intercept.

3x+y=15
Subtract 3x from both sides
y=-3x+15
Here slope, m=-3

Answer: Slope is -3. Option C is correct

Lilit [14]4 years ago

5 0

Slope intercept form is y = mx + b

so you must isolate the y first by moving the x (and anything attached to it) to the other side

3x (-3x) + y = (-3x) + 15

y = -3x + 15

The slope is m, which is next to x

-3x is what you are looking at

-3 is your slope

C. is your answer

hope this helps

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Please show ur work ty so much I appreciate it if u rlly help me

KengaRu [80]

The answer is going to be A!!!

8 0

3 years ago

Read 2 more answers

Give the following equations determine if the lines are parallel perpendicular or neither

GaryK [48]

In order to determine whether the equations are parallel, perpendicular, or neither, let's simply each equation into a slope-intercept form or basically, solve for y.

Let's start with the first equation.

$\frac{6x-5y}{2}=x+1$

Cross multiply both sides of the equation.

$6x-5y=2(x+1)$ $6x-5y=2x+2$

Subtract 6x on both sides of the equation.

$6x-5y-6x=2x+2-6x$ $-5y=-4x+2$

Divide both sides of the equation by -5.

$-\frac{5y}{-5}=\frac{-4x}{-5}+\frac{2}{-5}$ $y=\frac{4}{5}x-\frac{2}{5}$

Therefore, the slope of the first equation is 4/5.

Let's now simplify the second equation.

$-4y-x=4x+5$

Add x on both sides of the equation.

$-4y-x+x=4x+5+x$ $-4y=5x+5$

Divide both sides of the equation by -4.

$\frac{-4y}{-4}=\frac{5x}{-4}+\frac{5}{-4}$ $y=-\frac{5}{4}x-\frac{5}{4}$

Therefore, the slope of the second equation is -5/4.

Since the slope of each equation is the negative reciprocal of each other, then the graph of the two equations is perpendicular to each other.

5 0

1 year ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Tomtit [17]

Apparently my answer was unclear the first time?

The flux of F across S is given by the surface integral,

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S$

Parameterize S by the vector-valued function r(u, v) defined by

$\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2. Then the surface element is

dS = n • dS

where n is the normal vector to the surface. Take it to be

$\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}$

The surface element reduces to

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv$

so that it points toward the origin at any point on S.

Then the integral with respect to u and v is

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS$

$=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS$

$=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}$

4 0

4 years ago

What is an equivalent expression for 2x + 10x and 2(x-4) + x

tester [92]

2x+10x combine like terms
12x
2(x-4)+4 Distribute 2
2x-8+4
combine like terms
2x-4

6 0

3 years ago

The stack on the left is made up of 15 pennies, and the stack on the right is also made up of 15 pennies. If the volume of the s

valkas [14]

The volumes are the same.

4 0

3 years ago