Answer:
-1.5
Step-by-step explanation:
Answer:-25
Step-by-step explanation:
simplify
-1 < c + 2 < 3....subtract 2 from all sections
-1 - 2 < c + 2 - 2 < 3 - 2...simplify
-3 < c < 1
==============================
32 > 16 - 4g > 12....subtract 16 from all sections
32 - 16 > 16 - 16 - 4g > 12 - 16....simplify
16 > -4g > -4 ...now divide all sections by -4, and change inequality signs
16/-4 < (-4/-4)g < -4/-4...simplify
-4 < g < 1
============================
6y + 1 > = 10
6y > = 10 - 1
6y > = 9
y > = 9/6 which reduces to 3/2 or 1 1/2
-3/2y > = 9 ....multiply both sides by -2/3, cancelling the -3/2 on the left...and change the inequality sign
y < = 9 * -2/3
y < = -18/3 which reduces to - 6
so y > = 9/6(or 1 1/2) or y < = -6
The answer is 640 because if you multiply the total student of 6h grade by 100 and divide by 30. You will get the total of student in the school
By the fundamental theorem of algebra, we can write
![x^2+bx+10=(x-r_1)(x-r_2)](https://tex.z-dn.net/?f=x%5E2%2Bbx%2B10%3D%28x-r_1%29%28x-r_2%29)
Expanding the right hand side, we get
![x^2+bx+10=x^2-(r_1+r_2)x+r_1r_2\implies\begin{cases}r_1+r_2=-b\\r_1r_2=10\end{cases}](https://tex.z-dn.net/?f=x%5E2%2Bbx%2B10%3Dx%5E2-%28r_1%2Br_2%29x%2Br_1r_2%5Cimplies%5Cbegin%7Bcases%7Dr_1%2Br_2%3D-b%5C%5Cr_1r_2%3D10%5Cend%7Bcases%7D)
We want
to be an integer, which means
must also be integers. This means
must be factors of 10. There are several possibilities:
![r_1=\pm10,r_2=\pm1\implies b=-(10+1)=-11\text{ or }b=-(-10-1)=11](https://tex.z-dn.net/?f=r_1%3D%5Cpm10%2Cr_2%3D%5Cpm1%5Cimplies%20b%3D-%2810%2B1%29%3D-11%5Ctext%7B%20or%20%7Db%3D-%28-10-1%29%3D11)
![r_1=\pm5,r_2=\pm2\implies b=-(5+2)=-7\text{ or }b=-(-5-2)=7](https://tex.z-dn.net/?f=r_1%3D%5Cpm5%2Cr_2%3D%5Cpm2%5Cimplies%20b%3D-%285%2B2%29%3D-7%5Ctext%7B%20or%20%7Db%3D-%28-5-2%29%3D7)
So there are 4 possible values for
: -11, -7, 7, and 11.