Unit 6 lesson 4 practice problems

{(-3.7).(-5, -8), (-6, -1). (-9, -1), (0, -3) }

zhenek [66]

Answer:

The domain is the set of all the values of x, so it would be -3, -5, -6, -9, 0. The range would be every set of y, so it would be 7, -8, -1, -3. Finally, this relation is indeed a function.

Domain: -3, -5, -6, -9, 0

Range: 7, -8, -1, -3

Is it a function? Yes

4 0

2 years ago

Wegnerkolmp or someone please help me with this question about slope....

NNADVOKAT [17]

Answer:

The slope is -3/4

Step-by-step explanation:

We need two points to find the slope

We have one point at (0,5) and we have one point at (4,2)

We can use the slope formula

m = (y2-y2)/(x2-x1)

= (2-5)/( 4-0)

= -3/4

The slope is -3/4

3 0

3 years ago

Read 2 more answers

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

2 years ago

In triangle ABC, the size of angle B is 5 times the size of angle A, and the size of angle C is 7 less than 4 times the size of

kipiarov [429]

It sounds like everything is being described in reference to angle A, so a good starting point is to pick a variable to represent the measure of angle A. I'm going to use "a" for that.

Next, I'm going to take the verbal descriptions in the problem and "translate" them into "math language". They say angle B is 5 times angle A, so that means the measure of angle B is actually 5a. The size of angle C is 5 degrees less than 4 times the size of A, so that translates to 4a-5.

We now know the following:

Angle A: a
Angle B: 5a
Angle C: 4a-5

Now, to find the value of a, we need to remember that all the angles in a triangle add up to 180 degrees.

So we have: a+5a+4a-5 = 180

We can solve that equation by combining like terms to get: 10a-5=180

We can add 5 to both sides: 10a =185

And divide by 10: a = 18.5

That tells us the measure of angle A! (It's 18.5 degrees). Now we can go back and find B by multiplying A by 5. We get 92.5 degrees for that.

Finally, we can find C by taking 4*18.5-5. We get 69 degrees for that.

One last thing-- let's check that they really do add up to 180! 69+92.5+18.5 = 180. Yep!

Hope that helps you!!

5 0

2 years ago

PLEASE HELP I DONT KNOW WHAT TO DO !! I APPRECIATE IT IF YOU HELP ME !

attashe74 [19]

Answer:

x = ±20

Step-by-step explanation:

x^2 -1=399

Add 1 to each side

x^2 = 399+1

x^2 = 400

Take the square root of each side

sqrt(x^2) = ±sqrt(400)

x = ±20

8 0

2 years ago