2 is 1/10 of what number

What is the image of the point (-3,-6) on rotation of 90 degrees about the origin ?

SOVA2 [1]

Answer:

(- 6, 3 )

Step-by-step explanation:

Under a rotation about the origin of 90°

a point (x, y ) → (y, - x ), thus

(- 3, - 6 ) → (- 6, 3 )

6 0

3 years ago

Please help me with these questions

Sindrei [870]

1) 8^-24 = 1 / 8^24
2) 6^-1 = 1/6
3) 5^-8 = 1 / 5^8
4) -3^1 = -3
5) 4^-4 = 1 / 4^4
6) 5^9
7) 8^-3 = 1 / 8^3
8) 9^4
9) 7^1 = 7
10) 2^-4 = 1 / 2^4

6 0

3 years ago

How many degrees does the hour hand move from 3:00 to 5:30?

hodyreva [135]

now, keeping in mind that a circular clock has 360° total, and there are 12 hours labeled on it evenly, so the angle between each hour label is 360 ÷ 12 = 30.

so, from 3pm to 5pm, it moved by 30° + 30° = 60°.

from 5pm to 5:30, that's only half an hour, so it moved half 30°, or 15°.

60° + 15° = 75°.

4 0

3 years ago

PLEASE HELP AND HURRY

Naily [24]

Answer:

$42.79

Step-by-step explanation:

Tax: 2.7993 or 2.80

39.99 + 2.80 = 42.79

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

3 years ago