Answer:
Step-by-step explanation:
\mathrm{Multiply\:the\:numerator\:and\:denominator\:by:}\:100
\mathrm{Multiply\:the\:quotient\:digit}\:\left(0\right)\:\mathrm{by\:the\:divisor}\:505
\mathrm{Subtract}\:0\:\mathrm{from}\:23
40,000 divided by 10 is 4,000
Y= 1x + 8
both is increasing by one
Answer:
y = 7 and x =3
Step-by-step explanation:
use elimination method
so we gonna eliminate y first
9x = 1 - 4y
-7x = -7 + 4y
7| 9x = 1 - 4y
9| -7x = -7 + 4y
63x = 7 - 28y
-63x = -63 + 36y
add eqtn 1 to eqtn2
you will get
0 = -56 + 8y
56 = -56 + 56 + 8y
56 = 8y
56÷8 = 8y÷ 8
7 = y
also eliminate x as we have done wth y
4| 9x = 1 - 4y
4| -7x = -7 + 4y
36x = 4 - 16y
-28x = -28 + 16y
add the two equations
u will get
8x = -24 + 0
8x = -24
8x÷8 = 24÷ 8
x = 3
Answer:
Step-by-step explanation:
The Universal Set, n(U)=2092
Let the number who take all three subjects, 
Note that in the Venn Diagram, we have subtracted from each of the intersection of two sets.
The next step is to determine the number of students who study only each of the courses.
![n(S\:only)=1232-[103-x+x+23-x]=1106+x\\n(F\: only)=879-[103-x+x+14-x]=762+x\\n(R\:only)=114-[23-x+x+14-x]=77+x](https://tex.z-dn.net/?f=n%28S%5C%3Aonly%29%3D1232-%5B103-x%2Bx%2B23-x%5D%3D1106%2Bx%5C%5Cn%28F%5C%3A%20only%29%3D879-%5B103-x%2Bx%2B14-x%5D%3D762%2Bx%5C%5Cn%28R%5C%3Aonly%29%3D114-%5B23-x%2Bx%2B14-x%5D%3D77%2Bx)
These values are substituted in the second Venn diagram
Adding up all the values
2092=[1106+x]+[103-x]+x+[23-x]+[762+x]+[14-x]+[77+x]
2092=2085+x
x=2092-2085
x=7
The number of students who have taken courses in all three subjects,
