3 years ago

How do you find the derivative of f(x)=cos(x)f(x)=cos(x) ?

Mathematics

1 answer:

Lyrx [107]3 years ago

3 0

$f(x)=\cos x\\
f'(x)=-\sin x$

You might be interested in

Simplify. 14a^2bc^4/21abc^6

kramer

14a^2bc^4/21abc^6 = 2a/3b^2

7 0

3 years ago

Read 2 more answers

What is 3 divided by 12 over 5

Lemur [1.5K]

I think it’s 1/5 simplified

3 0

3 years ago

How do i find the volume of the last one? density= 2.7g/cm^3 mass= 0.85g

Liula [17]

You do mass divided by density

4 0

3 years ago

Can someone please tell me how to show work for this so I can get an answer I din't understand this!!! :,(

pav-90 [236]

Answer:

100% I looked it up and red an article

3 0

3 years ago

Read 2 more answers

ΔEFG is located at E (0, 0), F (−7, 4), and G (0, 8). Which statement correctly classifies ΔEFG?

MariettaO [177]

Answer:

ΔEFG is an isosceles triangle.

Step-by-step explanation:

Given:

E (0, 0),

F (−7, 4),

G (0, 8)

ΔEFG

Solution:

Distance formula

Distance d = $\sqrt{(x_2-x_1)^2 +( y_2-y_1)^2$

Step 1: Finding the length of EF

By using distance formula,

$EF = \sqrt{(-7 - 0)^2 + (4-0)^2}$

$EF = \sqrt{(49) + (16)}$

$EF = \sqrt{(49) + (16)}\\EF = \sqrt{65}\\$

Step 2: Finding the length of FG

By using distance formula,

$FG = \sqrt{(0-(-7))^2+(8-4)^2}\\FG = \sqrt{(7)^2 +(4)^2}\\FG = \sqrt{49 +16}\\FG = \sqrt{65}$

Step 2: Finding the length of GE

$GE= \sqrt{(0-0)^2 + (0-8)^2}\\\\GE =\sqrt{(-8)^2}\\GE = \sqrt{64}\\GE = 8$

Thus we could see that the sides EF = FG

So it is a isosceles triangle.

6 0

3 years ago