Please help! Algebra 1 math.

1 answer:

4 0

A. The data in the table represent a function. y=x/2. This is seen as for every value of x in the table, y=x/2(half of it). You can show this further by trying this fact on each value.

B. For the original function, when x =8, y=x/2=4
For the function given in this part, when x=8, y=3(8)-10
y=24-10 y=14
This means the relation given in this part has a greater value when x=8

C. When x=80, f(x)=3(80)-10
=240-10
=230

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Solve <br>|-4b-8|+|-1-b^2|+2b^3. b= -2<br>show steps please .

evablogger [386]

Ok, first put in the -2 for each b. That gives:
|-4(-2)-8|+|-1(-(-2))^2|+2(-2)^3
Let's do each section.
The first section is |-4(-2)-8)|
-4 times -2 is 8, minus 8 is 0. The absolute value of 0 is still 0.
Now we move on to |-1(-(-2))^2)|
First we do exponents
-(-2) is 2, and 2^2 is 4. 4 times -1 is -4. The absolute value of -4 is 4
Now the last section, 2(-2)^3
Exponents first: (-2)^3 is -2 * -2 * -2, which is -8.
-8*2=-16.
0+4+(-16)=-12

7 0

3 years ago

Please help me out on this :(( struggling !

Yuki888 [10]

Answer:

answer is B

Step-by-step explanation:

step by step explanation

3 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.