All categories

2 years ago

Two motorcyclists, who start their journeys at the same instance from two different cities,

Mathematics

1 answer:

padilas [110]2 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

The percent of area associated with z = 1.4 is ____%

Solnce55 [7]

The answer is A. 38.49

7 0

4 years ago

Read 2 more answers

6th grade math help me pleaseeee...

iren2701 [21]

Answer:

-3.8

Step-by-step explanation:

u = 3.8

u × -1 = 3.8 × -1

-u = -3.8

3 0

3 years ago

A is the smallest two digit whole number; "b" is the sum of prime numbers up to 10. what is a b?

Natasha2012 [34]

Whole numbers are like 0,1,2,3,4,5...
the smallest one is 10

prime numbers
I do not accept 1 as a prime number, if you agree, go to AAAAAAAAA
if you do think it's a prime number, go to BBBBBBB

AAAAAAA
2,3,5,7
2+3+5+7=17
a+b=10+17=27

BBBBBBBBBBB
1,2,3,5,7
1+2+3+5+7=18
a+b=10+18=28

if you do not think a is a prime number, then the answer is 27
if you do think 1 is prime number then the answer is 28

5 0

3 years ago

Read 2 more answers

What is the degree measure of

Setler [38]

Why u ask me jk hold on

4 0

3 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

3 years ago