If the 0 and 2 are powers it would be 16^0=1, 2^2=4
Now we have 1+4*y-1.
Im not sure if you want parenthesis to make it 1+4(y-1) or not.
Solving for the first one it would simplify to 4y.
With the parenthesis it would be 4y-3.
Thats it simplified, if it is equal to something you would solve for y.
Answer: 9+23p ≤ 170
Step-by-step explanation:
9+23p≤170
23p ≤ 170-9
23p ≤ 161/23
p≤7
Answer:
13y= x + 59
Step-by-step explanation:
gradient= y² - y¹ / x²- x¹
=5 - 4 / 6- -7
= 1/ 13
<h3>equation of line:
<u>y=mx+c</u></h3>
x y
point (6,5)
<u>substitute</u>:
5= <u> </u><u>1</u><u> </u> (6) + c
13
c = 5 -<u> </u><u>6</u><u> </u>
13
= <u>5</u><u>9</u>
13
y=mx+c
13y= 1x + 59
From the information given in the question,
the only possible conclusion is:
If angle 1 is measured and angle 7 is measured,
the two measurements will be identical.
If angle 1 and angle 7 are related to a particular shape or drawing,
then additional conclusions may be possible, but we'd need to see
the shape or drawing.
Hello there!
I'm assuming since there is no question, that you want an explanation for composite functions.
Today, I want to introduce you to a very new way of looking at functions. Think of a function as a machine. I'll call this machine f. When you plug something into this machine, it is an x-value. The machine changes the x-value into a new value which is called a y-value. This is how a function works.
With composite functions though, things get a little bit tricky. To make f(g(x)), you need to plug in x into the g machine, and it will give you an output. (y-value) The next thing you do is take that y-value and plug it into the g machine. The g machine then gives you a new value. This value is f(g(x)).
Let's do an example together...
f(x)=3x and g(x)=x²+4
if we want f(g(x)), first plug in x to the g machine. when plugging in x to the g machine, we get x²+4 as given in the question.
Now we must plug in g(x) into the f machine. Since g(x) is x²+4, we just replace x with x²+4.
We get 3(x²+4)
This means that f(g(x))=3(x²+4)
NOTE: If you are seeking help with an actual question, please message me in the comments and I will assist you shortly!
I hope this helps!
Best wishes :)
