-3x + 6y + 5 = -7
<u> -5 -5</u>
-3x + 6y = -12
-3x + 3x + 6y = -12 + 3x
<u>6y</u> = <u>-12 + 3x</u>
6 6
y = -2 + 1/2x
-3x + 6(-2 + 1/2x) = -12
-3x - 12 + 3x = -12
-3x + 3x - 12 = -12
0x - 12 = -12
<u> +12 +12</u>
<u>0x</u> = <u>0</u>
0 0
x = 0
-3(0) + 6y = -12
0 + 6y = -12
<u>+0 +0</u>
<u>6y</u> = <u>-12</u>
6 6
y = -2
(x, y) = (0, -2)
Answer:
135 and 135
Step-by-step explanation:
The computation is shown below:
The number of examiners who passed in only one subject is as follows
= n(E) - n(E ∩M) + n(M) - n(E ∩M)
= (80 - 60 + 70 - 60)%
= 30%
Now the number of students who passed in minimum one subject is
n(E∪M) = n(E) + n(M) - n(E ∩M)
= 80 - + 70 - 60
= 90%
Now the number of students who failed in both subjects is
= 100 - 90%
= 10% of total students
= 45
So total number of students appeared for this 450
So, those who passed only one subject is
= 450 × 30%
= 135
Now the Number of students who failed in mathematics is
= 100% - Passed in Mathematics
= 100% - 70%
= 30% of 450
= 135
Answer:
Step-by-step explanation:
Factoring will reveal the solution. So we divide the equation by the greatest common factor of the terms and use that factor as the coefficient. In this case the greatest common factor is just x.
2x^2+5x
x(2x+5) so the equation will equal zero when either of those expressions is zero because zero times anything is zero. x=0 and x=-5/2
Step-by-step explanation:
b2 =5
.....................
1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok
2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No
3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No
4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok
5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No
Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10
