All categories

3 years ago

We're do I find a rate of change

Mathematics

2 answers:

SVEN [57.7K]3 years ago

6 0

You can find the rate of change by taking it's derivative.

andriy [413]3 years ago

4 0

On a graph you look on the slope and if there is a change in direction, that is the rate of change

You might be interested in

rancho middle school has an average of 23 3/8 students per teacher write this mixed number as a decimal

Reika [66]

let's firstly convert it to an improper fraction.

$\bf \stackrel{mixed}{23\frac{3}{8}}\implies \cfrac{23\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{187}{8}}\implies 187\div 8\implies \stackrel{decimal}{23.375}$

6 0

3 years ago

The answer to this equation

12345 [234]

Answer:

x=3

Step-by-step explanation:

We are given the equation

$log5x=log(2x+9)$

This can be rewritten as

$log5x-log(2x+9)=0$

Now using the quotient rule for logarithms we can combine these two

$log(\frac{5x}{2x+9} )=0$

Next we can remove the log by using an inverse operation

$10^{log(\frac{5x}{2x+9} )} =10^0\\\\\frac{5x}{2x+9}=1$

Now we can solve for x

$\frac{5x}{2x+9}=1\\\\5x=2x+9\\\\3x=9\\\\x=3$

4 0

3 years ago

if the measure of the third angle of the triangle is 20 more than three times the measure of either of the other two angles, fin

valentina_108 [34]

You should multiply 20 time 8 to get your answer

5 0

3 years ago

QUICK!! TEN POINTS!! BRAINLIEST FOR WHOEVER GETS IT RIGHT!!

lara [203]

Answer:

(-2, -4.5)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In the calm waters of an inlet, Manon's boat moves with a velocity (speed and direction) vector \vec{b_1} = (3,4) b 1 =(3,4)

andrew-mc [135]

Answer with Step-by-step explanation:

We are given that

$b_1=(3,4)$

$b_2=(1,1)$

We have to find the speed of river's current and the direction of current flowing.

$b=b_2-b_1=(1,1)-(3,4)=(-2,-3)$

Speed of river's current= $\mid b\mid$

Speed of river's current= $\sqrt{(-2)^2+(-3)^2}$

Speed of river's current= $\sqrt{13} m/s$

By using the formula

$\mid r\mid =\sqrt{x^2+y^2}$

Direction of current flowing, $\theta=tan^{-1}(\frac{y}{x})$

$\theta=tan^{-1}(\frac{-3}{-2})=56.3^{\circ}$

The angle lies in third quadrant therefore

$\theta=180+56.3=236.3^{\circ}$

3 0

3 years ago