Explain what a finite sequence might refer to.

Which point is on the origin?

g100num [7]

Ur answer is point c

4 0

3 years ago

Read 2 more answers

6^2r=800 round it to the nearest hundreth

kogti [31]

Question: solve for r: 6 ^ (2r) = 800

Answer: 1.87

Step-by-step explanation:

6 ^ (2r) = 800

log(6 ^ (2r)) = log(800)

2r log(6) = log(800)

r = log(800) / (2 log(6))

r = 1.86537641979

round

r = 1.87

log can be any logarithm function.

6^(2×1.86) = 784.734328546

6^(2×1.87) = 813.365367336

7 0

3 years ago

A coordinate plane graph is shown. Point C is at negative 4 comma 3. Point D is at 1 comma 0. A segment connects the two points.

lina2011 [118]

The distance between them is 5.83 units. Enjoy!

5 0

4 years ago

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A line segment or ray that is perpendicular to the segment at its midpoint is called

Dimas [21]

Perpendicular bisector

3 0

3 years ago

Find the directional derivative of the function at the given point in the direction of the vector v. h(r, s, t) = ln(3r + 6s + 9

vodka [1.7K]

Answer:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=(\frac{2}{7}\overrightarrow{\rm i})*(\frac{1}{r+2s+3t})+(\frac{6}{7}\overrightarrow{\rm j})*(\frac{2}{r+2s+3t})+(\frac{3}{7}\overrightarrow{\rm k})*(\frac{3}{r+2s+3t})$

Step-by-step explanation:

First, let’s check to see if the direction vector is a unit vector:

$||v||=\sqrt{(14)^{2} +(42)^{2} +(21)^{2} } =\sqrt{2401} =49$

It’s not a unit vector. therefore let's divide the vector by its magnitude in order to convert it into a unit vector:

$\overrightarrow{\rm v}=(\frac{14}{49}\overrightarrow{\rm i})+(\frac{42}{49}\overrightarrow{\rm j})+(\frac{21}{49}\overrightarrow{\rm k})= (\frac{2}{7}\overrightarrow{\rm i})+(\frac{6}{7}\overrightarrow{\rm j})+(\frac{3}{7}\overrightarrow{\rm k})$

Now, the directional derivative is given by:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=\frac{\partial h(r,s,t)}{\partial r}\overrightarrow{\rm i} + \frac{\partial h(r,s,t)}{\partial s}\overrightarrow{\rm j} + \frac{\partial h(r,s,t)}{\partial t}\overrightarrow{\rm k}$

So let's calculate the partial derivates:

$\frac{\partial }{\partial r} ln(3r+6s+9t)=\frac{3}{3r+6s+9t}=\frac{1}{r+2s+3t}$

$\frac{\partial }{\partial s} ln(3r+6s+9t)=\frac{6}{3r+6s+9t}=\frac{2}{r+2s+3t}$

$\frac{\partial }{\partial t} ln(3r+6s+9t)=\frac{9}{3r+6s+9t}=\frac{3}{r+2s+3t}$

Therefore:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=(\frac{2}{7}\overrightarrow{\rm i})*(\frac{1}{r+2s+3t})+(\frac{6}{7}\overrightarrow{\rm j})*(\frac{2}{r+2s+3t})+(\frac{3}{7}\overrightarrow{\rm k})*(\frac{3}{r+2s+3t})$

3 0

4 years ago