Answer:
0.35m + 25 = 42.50
Step-by-step explanation:
If you were to rent a truck for one day you will first have to pay a $25 fee for one day. This is represented by + 25 in the equation above.
Then for every mile you go you must pay $0.35, represented as 0.35m.
<em>As an example to make it easier if you had to pay 1 dollar per mile and you went 10 miles it would cost $10. This can be proved by 1m (m being the variable for number of miles) Plug in 10 for m and you can asee 1(10)=10, because obviously 1 times 10 is 10. </em>
Same premise with $0.35
Then both of those fees added up are equal to a total of $42.50.
To prove it can find the exact number of miles that need to be driven to cost a total of $42.50 just follow steps bellow.
1. 0.35m + 25 - 25 = 42.50 - 25
Subtracting the inital fee since it is part of the total, and if you add/subtract/multiply/divide on one side you must do to the other.
2. 0.35m/0.35 = 17.50/0.35
You divide because m is being multiplied with 0.35 and to undo the multiplication you divide. Doing the same to both sides.
3. m = 50
Plug in 50 to "m" in the equation above and if both sides are equal the answer is valid.
Answer:
125 degrees.
Step-by-step explanation:
L is the midpoint of NK so the arcs NM and Mk are equal.
So MK = 55 degrees and the arcs JN and JK are equal
Therefore
JK =
(360 - 2*55)/2
= (360 - 110) / 2
= 250 /2
= 125 degrees.
80 ( i think )
sorry if it's wrong but im pretty sure that is it
For example, for A and A'
Let:
R=(0,0) the reference point:
![\begin{gathered} \text{Distance between A and R:} \\ d=\sqrt[]{(4-0)^2+(2-0)^2}=2\sqrt[]{5} \\ \text{Distance between A' and R} \\ d=\sqrt[]{(4-0)^2+(-2-0)^2}=2\sqrt[]{5} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7BDistance%20between%20A%20and%20R%3A%7D%20%5C%5C%20d%3D%5Csqrt%5B%5D%7B%284-0%29%5E2%2B%282-0%29%5E2%7D%3D2%5Csqrt%5B%5D%7B5%7D%20%5C%5C%20%5Ctext%7BDistance%20between%20A%27%20and%20R%7D%20%5C%5C%20d%3D%5Csqrt%5B%5D%7B%284-0%29%5E2%2B%28-2-0%29%5E2%7D%3D2%5Csqrt%5B%5D%7B5%7D%20%5Cend%7Bgathered%7D)
