For this problem you are going to need to know the point-slope equation which is:
y-y₁=m(x-x₁)
Now, m=slope and the x and y values are given to us already, so now we just plug our variables into the formula.
It should look like this:

Since all they are asking for is for you to put it into a formula this will be your answer.
hope it helps u to get your abswer and me to be ranked as brainliest:)))
Answer:
A. 2^11
Step-by-step explanation:
(They are basically asking what's 2^4 × 2^7, but with more words.)
I usually do each exponent individually:
2^4 is the same as 2 × 2 × 2 × 2 = 16 (or you could have read the text to figure that out)
2^7 is the same as 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128
Then just multiply 128 and 16 to get 2,048, and see which option also gives you 2,048.
BUT, you can also:
(Combine the exponents together to get your answer. Just remember that if it's multiplication you add them, and if it's division you subtract them.)
2^4 × 2^7
4 + 7 = 11
2^11 (This equals 2,048 btw. You don't even have to check all the options to get the answer).
Hope this helps friend :)
The last part I learned from another user, while answering one of your other questions. I personally find this mind blowing, lol.
Answer:
D)
Step-by-step explanation:
y = 12x.
A) x=8, so that y = 12(8) = 96. It's on the line
B) x = 10, so that y = 12(10) = 120. It's on the line
C) x = 15, so that y = 15(12) = 180. It's on the line
D) x = 18, so that y = 18(12) = 216, not equal to 206. So, the point is not on the line.
Answer:




Step-by-step explanation:
Given

Required
Solve (a) to (d)
Using tan formula, we have:

This gives:

Rewrite as:

Using a unit ratio;

Using Pythagoras theorem, we have:




Take square roots of both sides

So, we have:


Solving (a):

This is calculated as:


Where:


So:




Solving (b):

This is calculated as:

Where:
---- given
So:


Solving (c):

In trigonometry:

Hence:

Solving (d):

This is calculated as:


Where:


So:


