Step-by-step explanation:

<u>Answer:</u>
C. Left-handed: 48, Right-handed: 504
D. Left-handed: 30, Right-handed: 315
<u>Step-by-step explanation:</u>
We are given that there are 58 left-handed students and 609 right-handed students at East Middle School and these numbers of students are proportional to the number of left and right handed students at East Middle School.
Given the above information, we are are to determine which two options could be the the numbers of left-handed and right-handed students at West Junior High.
Ratio of right handed to left handed students at East Middle School = 
Checking for ratios of the given options:
A. 
B. 
C. 
D. 
E. 
Therefore, the possible numbers of left-handed and right-handed students at West Junior High could be C. Left-handed: 48, Right-handed: 504 and D. Left-handed: 30, Right-handed: 315.
Step-by-step explanation:

Given equaiton is in the form of ax^2 +bx+c=0
we apply quadratic formula to solve for x

a= 1 b = -12 and c= 59



Divide the 12 and square root terms by 2

so
and 
Answer:
124.67cm³
Step-by-step explanation:
==>Given:
Dimensions of current can:
Height (h) = 12cm
Diameter = 6cm (radius = 3cm)
Volume of current can (V1) = 339.12 cm³
Dimensions of future cans to be increased multiple of 1.11:
height = 12cm × 1.11 = 13.32 cm
radius = 3cm × 1.11 = 3.33 cm
Volume of future can (V2) = πr²h = 3.14*3.33²*13.32
= 3.14*11.0889*13.32 = 463.791025
V2 ≈ 463.79 cm³
==>Find how much more would new cans hold = Volume of new can - volume of current can
= 463.79 cm³ - 339.12 cm³
= 124.67cm³
Do u mean from the book or something?