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
The answer is 13
If R=4 the the equation would be 4+9 which equals 13.
Answer:
(x-3)(x^2-7) = x^3 -3x^2 -7x + 21
just change the sign of the root and put x in front of it, then multiply the factors all together
(x-3)(x-sqr7)(x+sqr7) roots come in conjugate pairs to eliminate irrational coefficients
(x-sqr7)(x+sqr7) = x^2-7. similar to (a-b)(a+b) = a^2-b^2
Step-by-step explanation:
Step-by-step explanation:
You should prolly subtract 2 from both sides first.
So you get, -|2/5x + 3| > -1 2/5
- Add three to both sides
- -|2/5x| = -1 2/5
- Do -1 and 2/5 divided by 2/5 to get -3.5, x > -3.5
One of the answers is x > -3.5 but you need to switch the sign to less than and make the other side the opposite.
You see that, switched the sign and made whats on the right side opposite. So now solve for x again but with this new equation. Also, the parentheses are absolute value signs
Subtract 3 from both sides
-|2/5x| < -1 and 3/5
Divide 2/5 from each side to get x < -4
I hope I'm correct, my bad if I'm wrong
