The partial quotient is 23
Hey there!
![\left[\begin{array}{ccc}\boxed{\boxed{ \frac{6}{4}}} \end{array}\right] = \left[\begin{array}{ccc}\boxed{\boxed{150}}\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B%20%5Cfrac%7B6%7D%7B4%7D%7D%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B150%7D%7D%5Cend%7Barray%7D%5Cright%5D%20)
All I did was to multiply each side by 100 which then go me this!
Hope this helps!
Answer:
∠2 and ∠5
Step-by-step explanation:
we know that
<u>Alternate Exterior Angles</u> are a pair of angles on the outer side of each of those two lines but on opposite sides of the transversal
In this problem
∠12 and ∠2 are alternate exterior angles
∠12 and ∠5 are alternate exterior angles
therefore
∠2 and ∠5 are each separately alternate exterior angles with ∠12
Answer:
x y
5 20
15 10
45 5
Step-by-step explanation:
Pattern x:
Start with 5.
Multiply by 3: 5 * 3 = 15
Multiply by 3: 15 * 3 = 45
Pattern x:
Start with 20.
Multiply by 1/2: 20 * 1/2 = 10
Multiply by 1/2: 10 * 1/2 = 5
x y
5 20
15 10
45 5
Answer:
179 + 49+ 139 = 367
Step-by-step explanation:
