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3 years ago

Multiply (x-3)(4x+2) using the distributive property.

1 answer:

3 0

$\text{The distributive property:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(x-3)(4x+2)=(x)(4x)+(x)(2)-(3)(4x)-(3)(2)\\\\=4x^2+2x-12x-6=4x^2-10x-6$

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I need help please!!!!

zaharov [31]

Answer:

Step-by-step explanation:

The equations are all in point-slope form with a slope of 6/5. The points used are (-1, -2), (-2, -1), and (1, 2). It seems point (1, 2) best matches a point on the graphed line. Choice C is the best. (In the attached graph, choice C is the red line.)

_____

<em>Additional comment</em>

The point-slope form of the equation for a line is ...

y -k = m(x -h) . . . . . . . line with slope m through point (h, k)

The equation might be easier to see if the point chosen were one at a grid intersection, such as (-4, -4) or (6, 8).

8 0

2 years ago

16. Tell whether the lines for each pair of equations are parallel, perpendicular, or neither.

il63 [147K]

Answer:

Given equation are:

$y = -\frac{1}{4}x+8$ ......[1]

$-2x+8y = 4$ .....[2]

The two lines are parallel lines then their slopes will be equal.

When two lines are perpendicular then, the slope of lines are the negative reciprocals of each other.

Now, Equation of a line is in the form of y =mx+b where m is the slope of the line.

Slope $(m_1)$ of equation of line in [1] is;

$y = -\frac{1}{4}x+8$

then;

$m_1= -\frac{1}{4}$

Slope $(m_2)$ of equation of line in [2];

$-2x+8y = 4$

Add both sides 2x we get;

-2x + 8y + 2x = 2x + 4

Simplify:

8y = 2x +4

Divide both sides by 8 we get;

$y = \frac{1}{4} x + \frac{1}{2}$

then;

$m_2 = \frac{1}{4}$

Therefore, the given two lines are neither parallel nor perpendicular.

4 0

3 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$