We have been given two points.
and
. We are asked to find the point B such that it divides line segment AC so that the ratio of AB to BC is 4:1.
We will use segment formula to solve our given problem.
When a point P divides segment any segment internally in the ratio
, then coordinates of point P are:
![[\right x=\frac{mx_2+nx_1}{m+n},y=\frac{my_2+ny_1}{m+n}\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D%5Cfrac%7Bmx_2%2Bnx_1%7D%7Bm%2Bn%7D%2Cy%3D%5Cfrac%7Bmy_2%2Bny_1%7D%7Bm%2Bn%7D%5Cleft%5D)
and
.

Upon substituting our given information in above formula, we will get:
![[\right x=\frac{4(3)+1(3)}{4+1},y=\frac{4(9)+1(4)}{4+1}\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D%5Cfrac%7B4%283%29%2B1%283%29%7D%7B4%2B1%7D%2Cy%3D%5Cfrac%7B4%289%29%2B1%284%29%7D%7B4%2B1%7D%5Cleft%5D)
![[\right x=\frac{12+3}{5},y=\frac{36+4}{5}\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D%5Cfrac%7B12%2B3%7D%7B5%7D%2Cy%3D%5Cfrac%7B36%2B4%7D%7B5%7D%5Cleft%5D)
![[\right x=\frac{15}{5},y=\frac{40}{5}\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D%5Cfrac%7B15%7D%7B5%7D%2Cy%3D%5Cfrac%7B40%7D%7B5%7D%5Cleft%5D)
![[\right x=3,y=8\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D3%2Cy%3D8%5Cleft%5D)
Therefore, the coordinates of point B would be
.
Answer:
d. 15
Step-by-step explanation:
Putting the values in the shift 2 function
X1 + X2 ≥ 15
where x1= 13, and x2=2
13+12≥ 15
15≥ 15
At least 15 workers must be assigned to the shift 2.
The LP model questions require that the constraints are satisfied.
The constraint for the shift 2 is that the number of workers must be equal or greater than 15
This can be solved using other constraint functions e.g
Putting X4= 0 in
X1 + X4 ≥ 12
gives
X1 ≥ 12
Now Putting the value X1 ≥ 12 in shift 2 constraint
X1 + X2 ≥ 15
12+ 2≥ 15
14 ≥ 15
this does not satisfy the condition so this is wrong.
Now from
X2 + X3 ≥ 16
Putting X3= 14
X2 + 14 ≥ 16
gives
X2 ≥ 2
Putting these in the shift 2
X1 + X2 ≥ 15
13+2 ≥ 15
15 ≥ 15
Which gives the same result as above.
<h2>⚘Your Answer:------</h2>
<h3><u>Given</u><u> </u><u>Information</u><u>:</u></h3>
- <u>Diameter of can</u>:- 3.6 cm
- <u>Height of</u><u> </u><u>can</u><u>:</u><u>-</u> 6.4 cm
<h3><u>To</u><u> </u><u>Find</u><u> </u><u>Out</u><u>:</u></h3>
- <u>Volume</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>Can</u><u>.</u>
<h3><u>Solution</u><u>:</u></h3>
Radius = (Diameter/2)
ㅤㅤㅤㅤ3.6 cm/2
ㅤㅤㅤㅤ1.8 cm
<u>So</u><u>,</u><u> </u><u>Radius</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>soup</u><u> </u><u>can</u><u> </u><u>is</u> 1.8 cm.
Volume øf cylinder = Volume øf can
Volume of cylinder = π r²h
(π = 22/7)
(r = radius = 1.8 cm)
(h = height = 6.4 cm
ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= 22/7×(1.8 cm)²× 6.4 cm
ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= 22/7× 3.24 cm² × 6.4 cm
ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= 22/7 × 20.736 cm³
ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= (456.192/7) cm³
ㅤㅤㅤㅤㅤㅤㅤㅤㅤ= 65.17 cm³
So, the Volume of the soup can is 65.17 cm³
ㅤㅤㅤㅤㅤㅤㅤㅤ⚘Thank You
The denominator is the one on the bottom so 9 is the denominator the top is the numerator