Answer:
The geometric mean of the measures of the line segments AD and DC is 60/13
Step-by-step explanation:
Geometric mean: BD² = AD×DC
BD = √(AD×DC)
hypotenuse/leg = leg/part
ΔADB: AC/12 = 12/AD
AC×AD = 12×12 = 144
AD = 144/AC
ΔBDC: AC/5 = 5/DC
AC×DC = 5×5 = 25
DC = 25/AC
BD = √[(144/AC)(25/AC)]
BD = (12×5)/AC
BD= 60/AC
Apply Pythagoras theorem in ΔABC
AC² = 12² + 5²
AC² = 144+ 25 = 169
AC = √169 = 13
BD = 60/13
The geometric mean of the measures of the line segments AD and DC is BD = 60/13
Answer:

General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
<u>Algebra I</u>
- Slope Formula:

Step-by-step explanation:
<u>Step 1: Define</u>
Point (4, 6)
Point (0, 8)
<u>Step 2: Find slope </u><em><u>m</u></em>
- Substitute:

- Subtract:

- Simplify:

Answer:
about 5.22hrs
Step-by-step explanation:
cross multiply
<u>120.45</u> <u>?</u>
15.67 40.15
120.45*40.15
=629.1505
629.1505/15.67
=5.2233333...
The total of 2 buckets of popcorn and 3 boxes of candy would be $23.25
To answer this question you need to form a set of simultaneous equations and solve them. We can do this by saying that a bucket of popcorn = P, and a box of candy = C. Then we can say:
4P + 6C = 46.50
P + C = 9.75
There are then two possible ways to solve; you can either say that C = 9.75 - P using the second equation and then substitute it into the first, or you can multiply the second equation by either 4 or 6 to cancel out P or C.
I’m going to multiply the second equation by 4:
4P + 4C = 39
Now we can subtract this for, the first equation:
4P + 6C = 46.50
4P + 4C = 39
2C = 7.50
C = 3.75
Now we can substitute this value of C into one of the equations to find P:
P + C = 9.75
P + 3.75 = 9.75
P = 6
And now to answer the question, you just multiply P by 2 and C by 3 and add them together, which gives you $23.25
I hope this helps! Let me know if you have any questions :)
Answer: 1 and 2/3
Step-by-step explanation:
Imagine 3 1/3 as if it’s in parenthesis as (3 + 1/3)
Subtract three from five.
You should get two.
Then you subtract the last 1/3 from that two.
There are three thirds in one digit.
Therefore, two thirds are left over, plus that last whole digit.
You are left with one and one third.