45 POINTS

The measure of an angle is 29.3°. What is the measure of a complementary angle?

crimeas [40]

The answer is 60.7 because 90-29.3=60.7

7 0

2 years ago

Two squares have areas of 4 cm?

timofeeve [1]

Answer:

1/2

Step-by-step explanation:

First square area = 4cm^2, thus the side length is 2cm

Second square area = 16cm^2, thus the side length is 4cm

This gives a ratio of 2/4, simplifying yields 1/2

3 0

1 year ago

I need help with -1 + (-4) + (-9) PLEASE DO THE STEP BY STEP EXPLANATION

Lera25 [3.4K]

Answer:

yes

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Please help me asap!!!

luda_lava [24]

Answer:

Choice A.......< B is 147 degrees

Step-by-step explanation:

The total sum of the interior angles of a pentagon is 540 degrees

...So let's count the angles that we already have:

<A is 90

<B is ?

<C is 105

<D is 108

<E is 90

Noticed that both angles A and E are 90 since they have that little square that represents a straight angle.

Let's find the value for <B

90+ x + 105 + 108 + 90= 540

393 + x= 540

x= 540 -393

x = 147

So angle B is 147 degrees

5 0

2 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

2 years ago