Through (-3,-3) and (-5,4)

Find the equation of the axis of symmetry.<br>equation of axis of symmetry:

sleet_krkn [62]

9514 1404 393

Answer:

x = -4

Step-by-step explanation:

The axis of symmetry is the vertical line through the vertex. The vertex is the turning point, or minimum, of the function. It is (x, y) = (-4, 0). The x-coordinate there is -4, so the equation of the axis of symmetry is ...

x = -4

5 0

3 years ago

Add the two expressions. 4.6x−3 and −5.3x+9 Enter your answer, in simplified form, in the box. PLEASE ANSWER FAST THIS IS A BIG

Ivan

To add the two expressions, we can write it as:

4.6x-3 + (-5.3x+9)

We can distribute the plus sign (which means just drop the parenthesis in this case):

4.6x-3-5.3x+9

Now, we can simplify by combining like terms:

-0.7x+6

4 0

3 years ago

Read 2 more answers

Can someone please solve this ASAP?

olga2289 [7]

The ladder and the wall form a right triangle.
The length is the hypotenuse = 15.
The base = 2*(15/5) = 2*3 = 6

Use pythagorean theorem to find other side of triangle:
$a^2 + b^2 = c^2 \\ \\ a^2 + 6^2 = 15^2 \\ \\ a^2 + 36 = 225 \\ \\ a^2 = 189 \\ \\ a = \sqrt{189} = 3 \sqrt{21}$

8 0

3 years ago

So who is gonna help me with this ? هههههههه please and thank you .

mario62 [17]

I believe you would multiply 6 and 10 and get the area of 60, thats all i can think of

8 0

3 years ago

Read 2 more answers

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