Answer:
c
Step-by-step explanation:
Answer:

Step-by-step explanation:
We want to find an expression that has a value of

First option:

Second option:

Correct option
Third option

Fourth option:

Step-by-step explanation:
It is an isosceles triangle , it has two equal angles both 55°.
All Isosceles triangles have two equal angles <em>a</em><em>n</em><em>d</em> two equal sides.
So in this case x +18 is equal to 3x, and to find x you will equal them to each other to create a linear equation and solve.
x + 18 = 3x
x - 3x = -18
-2x = -18
× = -18 / -2
× = 9 <em>(</em><em>a</em><em>n</em><em>s</em><em>)</em><em> </em>
Hope I was able to help. Good Luck :)))
Answer:
10010
Step-by-step explanation:


So
gives us:



-----------------------------------------------------
Combine like terms:


We aren't allowed to have a coefficient bigger than 1.
I'm going to replace
with 1 and
with
:

I want a
number:

Combine like terms:

:

Combine like terms:

We can rewrite the first term by law of exponents:


So the binary form is:

Maybe you like this way more:
Keep in mind 1+1=10 and that 1+1+1=11:
Setup:
1 0 1 1
+ 1 1 1
------------------------------
(1) (1) (1)
1 0 1 1
+ 1 1 1
------------------------------
1 0 0 1 0
I had to do some carry over with my 1+1=10 and 1+1+1=11.
Hi!
If she worked 18 hours last week and 20 hours this week, then she worked 38 hours in total, because 18 + 20 = 38.
If she earns $6 per hour, and she worked for 38 hours, then she got 38 sets of 6, which you can find the answer to by multiplying 38 and 6.
This is essentially 6 + 6 + 6 +...
38 * 6 = $228
So she earned $228 these two weeks.
Hope this helped!