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2 years ago

5

Now consider the family of lines ax + by = c such that a, b and c are consecutive multiples of each other. what would that mean?

give some examples of such lines in this family (coordinates.) what do you notice or wonder?

1 answer:

zhenek [66]2 years ago

7 0

Answer:

The first thing that was a good app is the app it needs a little bit of the app and it keeps crashing when it comes to my iPad to play with the iPhone and iPad version of my iPad to get my iPhone and iPhone app for my iPad to use it on the iPad app to get a iPhone app for my iPad iPad to get the iPhone version of my iPad version and I can get a few of my iPhone and iPhone version

Step-by-step explanation:

You might be interested in

-100/-20 multiplying integers

VLD [36.1K]

The answer is 20!

−100×(−20)=?
2000

4 0

2 years ago

Simplify the expression. A) 8^4•8^9

OlgaM077 [116]

Answer:

${8}^{13}$

Step-by-step explanation:

${8}^{4} \times {8}^{9} \\ {8}^{4 + 9} \\ {8}^{13} \\ = 5.497558139 \times 10 ^{11}$

Hope this helps you.

Let me know if you have any other questions :-)

7 0

3 years ago

Which of the following equations is true?

Korolek [52]

Answer:

B

Step-by-step explanation:

10×1=10

0+10=10

Hence they're equal

7 0

3 years ago

3. Find f[h(a+4)], given the functions below. *

mote1985 [20]

Answer:

C 5a^2 +70a +240

Step-by-step explanation:

We have given these following functions:

$f(x) = 5x, g(x) = -2x + 1, h(x) = x^2 + 6x + 8$

h(a+4)

This function is:

$h(a+4) = (a+4)^2 + 6(a + 4) + 8$

$h(a+4) = a^2 + 8a + 16 + 6a + 24 + 8$

$h(a+4) = a^2 + 14a + 48$

f[h(a+4)]

$f(x) = 5x$

Thus

$f(h(a+4)) = f(a^2+14a+48) = 5(a^2 + 14a + 48) = 5a^2 + 70a + 240$

The correct answer is given by option C.

8 0

3 years ago

Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line seg

n200080 [17]

$\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k$

is conservative if there is a scalar function $f(x,y,z)$ such that $\nabla f=\vec F$ . This would require

$\dfrac{\partial f}{\partial x}=y$

$\dfrac{\partial f}{\partial y}=x$

$\dfrac{\partial f}{\partial z}=3$

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

$f(x,y,z)=xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$f(x,y,z)=xy+h(z)$

$\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C$

$f(x,y,z)=xy+3z+C$

so $\vec F$ is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

$\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r$

$=f(5,7,-2)-f(1,2,3)=\boxed{18}$

8 0

3 years ago