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4 years ago

Find the slope-intercept form of an equation of the line perpendicular to the graph of x - 3y = 5 and passes through (0,6).

Mathematics

1 answer:

marissa [1.9K]4 years ago

7 0

Answer:

the desired equation is y = -3x + 6.

Step-by-step explanation:

1) Rewrite x - 3y = 5 in slope-intercept form: x - 5 = 3y, or y = (1/3)(x - 5)

2) Identify the slope of the given line. It is (1/3).

3) Find the slope of a line perpendicular to this one. It is the negative reciprocal of (1/3), or -3.

4) Use the slope-intercept form of the equation of a straight line, y = mx + b, to determine the b value and thus the equation of the perpendicular line:

6 = -3(0) + b. Then b = 6, and the desired equation is y = -3x + 6.

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4 years ago

Find f(-5) f(x) -4x + 3

Fiesta28 [93]

Answer:

if f(x)= -4x+3

f(-5)= 23

Step-by-step explanation:

first you substitute x for -5

f(-5)= -4(-5)+3

then you multiply

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then you just add the rest (20+3) which will give you your answer

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3 years ago

The point slope form of a line with a slope of 2 through (1, 4) is y – 4 = 2(x – 1).

ella [17]

Answer:

TRUE

Step-by-step explanation:

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And the slope is 2.

8 0

3 years ago

Please help me out! i have a few more of these questions, about 4

Eva8 [605]

Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

3 years ago