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3 years ago

Use the given conditions to write an equation for each line in general form.

Mathematics

1 answer:

Aneli [31]3 years ago

6 0

Answer:

It's the last choice:

2x + y - 1 = 0.

Step-by-step explanation:

x - 2y - 3 = 0

Convert to slope-intercept form in order to find the slope:

-2y = -x + 3

y = 1/2x - 3/2

So the slope is 1/2. The slope of a line perpendicular to this is -1 / 1/2 = -2.

Using the point-slope form to find the required equation:

y - y1 = m(x - x1)

x = 4 and y = -7 so we have:

y + 7 = -2(x - 4)

y + 7 = -2x + 8

2x + y - 1 = 0 is the answer.

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Find the measure of each angle and tell which postulates or theorems you used <br> ?

Illusion [34]

Answer:

∠6=78° ∠7=102° ∠14=78°

Step-by-step explanation:

Angles 6 and 7 are supplementary angles so ∠a+∠b=180°

in this case it would be 102°+x=180° and you subtract 102° from 180° getting 78°

∠7 is the vertical angles postulate stating that any angles vertical to each other are eaqual

7 0

3 years ago

Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector

Feliz [49]

$f(x,y,z)=2z^2x+y^3$

$f$ has gradient

$\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k$

which at the point (-1, 4, 3) has a value of

$\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k$

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say $\vec u=15\,\vec\imath+25\,\vec\jmath$ , in which case we have

$\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}$

Then the derivative of $f$ at (-1, 4, 3) in the direction of $\vec u$ is

$D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}$

4 0

3 years ago

Hhhhhhhellllllllp.....................................

Anit [1.1K]

You have 28 meters of fence to make 3 sides.
28/3 = 9.3333

The pen will be 9.3333 by 9.3333 meters

6 0

3 years ago

(35 points) answer 5 and 7 only please

tatyana61 [14]

A2+b2=c2 and get your answer

3 0

3 years ago

If ( f ∘ g)(x) = x2 - 6x + 8 and g(x) = x - 3, what is f(x)?

DochEvi [55]

$\bf \begin{cases}
(f\circ g)(x)=x^2-6x+8\\\\
(f\circ g)(x)=f(~~g(x)~~)\\\\
g(x)=x-3\\
\end{cases}
\\\\\\
\textit{now, one probability is }x^2-1
\\\\\\
f(x)=x^2-1\implies f(~~g(x)~~)=(x-3)^2-1
\\\\\\
f(~~g(x)~~)=(x^2-6x+9)-1\implies f(~~g(x)~~)=x^2-6x+8$

5 0

3 years ago