Answers:
4; 20; 3x² - 4x + 3; 52; 17
Step-by-step explanation:
f(-1): replace x in f(x) = x² + 3 with -1: f(-1) = (-1)² + 3 = 4
f(-4) + g(-1) = (-4)² + 3 + <em>2(-1) + 3</em> = 16 + 3 <em>- 2 + 3</em> = 20 <em>(since g(x) = 2X + 3)</em>
<em />
3f(x) - 2g(x) = 3[x² +3] - 2[2x + 3} = 3x² + 9 - 4x - 6 = 3x² - 4x + 3
f(g(2)): First, evaluate g(2). It is g(2) = 2(2) + 3 = 7. Next, use this output, 7, as the input to f(x): f(g(x)) = (7)² + 3 = 49 + 3 = 52
g(f(2)): First, evaluate f(x) at x = 2: f(2) = (2)² + 3 = 7. Next, use this 7 as the input to g(x): g(f(2)) = g(7) = 2(7) + 3 = 17
<h2>Steps:</h2>
So for this, I will be completing the square to solve for x. Firstly, add 10 and subtract 30x on both sides of the equation:
![x^2-30x=10](https://tex.z-dn.net/?f=x%5E2-30x%3D10)
Next, we want to make the left side of the equation a perfect square. To find the constant of this soon-to-be perfect square, divide the x coefficient by 2 and square the quotient. Once you get that result, add it to both sides of the equation:
![30\div 2 = 15\\15^2=225\\\\x^2-30x+225=235](https://tex.z-dn.net/?f=30%5Cdiv%202%20%3D%2015%5C%5C15%5E2%3D225%5C%5C%5C%5Cx%5E2-30x%2B225%3D235)
Now, factor the left side:
![(x-15)^2=235](https://tex.z-dn.net/?f=%28x-15%29%5E2%3D235)
Next, square root both sides:
![x-15=\pm\ \sqrt{235}](https://tex.z-dn.net/?f=x-15%3D%5Cpm%5C%20%5Csqrt%7B235%7D)
Now, add 15 to both sides of the equation:
![x=15\pm \sqrt{235}](https://tex.z-dn.net/?f=x%3D15%5Cpm%20%5Csqrt%7B235%7D)
This is the <em>exact</em> solution. To find the approximate solution, solve the left side twice -- once with the plus sign, once with the minus sign:
![x=30.33,-0.33](https://tex.z-dn.net/?f=x%3D30.33%2C-0.33)
<h2>Answer:</h2>
In short:
- Exact Solution:
![x=15\pm\sqrt{235}](https://tex.z-dn.net/?f=x%3D15%5Cpm%5Csqrt%7B235%7D)
- Approximate Solution (Rounded to the hundredths):
![x=30.33,-0.33](https://tex.z-dn.net/?f=x%3D30.33%2C-0.33)
Answer:
-2x+19
Step-by-step explanation:
We open the parentheses, so
-12x+15+10x+4
Simplify
-2x+19
First off, you can rewrite your equation like so: c=(a-b)x. You can then plug in your given constraints for your choices. If a-b = 1 and c = 0 leaving: 0=1(x), x must equal 0 and only 0 as any constant multiplied by 0 equals 0. So that choice is eliminated. Now let's consider when a=b and c != 0. Since we are given a-b and a=b and c != 0, we have:
c = 0x. This contradicts our claim we made about our constraints. C cannot equal zero but we have a-b=0. Therefore, this claim makes no sense as any value for x will not satisfy the equation. This choice is valid. When a=b and c=0, we have: 0 = 0x. Here, x can be any value and still return 0 as an answer. This choice is valid. If a-b=1 and c != 1, we have: c = 1x. Our only rule here is that c cannot equal 1. This means that x can be any value other than 1 so this choice can be marked down. If a != b and c=0, this gives: 0 = (a-b)x. Given that a-b can be any value, x must be equal to only 0 to satisfy this equation so this choice can't be correct. So the right answers are: option 2, option 3, option 4 and option 5.
-2/3x+9=4/3x-3
+3 +3
-2/3x+12=4/3x
+2/3x +2/3x
12=6/3x
12=2x
x=6
x=6