He first two are relatively easy. f(x-2) shifts it sideways (2 to the right in this case) f(x) -2 shifts it straight down (by 2)
f(2x) means replace x with 2x in the original 4(2x + 1)^2 − 3 if you factor out the 2 from the square root you get 4*sqr(2) ( x + ½)^2 - 3 which is now in the form a ( x -h)^2 + k the vertex is now at (-½, -3) (it used to be at (-1,-3) the "a" has gotten bigger, which makes the parabola "skinnier" (it goes up faster)
2•f(x) becomes 2*( 4(x + 1)2 − 3 ) = 8 (x+1)^2 -6 where is the vertex now? is this parabola fatter or skinnier than the original f(x) ?
Answer:
3
Step-by-step explanation:
8*3 is 24 and you cannot add another 8 after that
Answer: No Solution
<u>Step-by-step explanation:</u>
ln x - ln (x + 2) = 4 restrictions: x > 0 and x + 2 > 0 → x > 0
ln
= 4
= e⁴
x = e⁴ (x + 2)
x = 54.5982 (x + 2)
x = 54.5982x + 109.1964
- 53.5982x = 109.1964
x = 
x = -2.0373
-2.0373 is not greater than 0 so is not valid
**************************************************************************
Answer: 2.1972
<u>Step-by-step explanation:</u>
eˣ = 9
ln eˣ = ln 9
x = ln 9
x ≈ 2.1972
**************************************************************************
Answer: 
<u>Step-by-step explanation:</u>
log₂ (3x - 4) = -1
3x - 4 = 2⁻¹
3x - 4 = 
<u> +4 </u> <u>+4 </u>
3x = 
3x = 

x = 
**************************************************************************
Answer: 1.117519
<u>Step-by-step explanation:</u>
in the calculator, type in 1 ÷ 9.
Then hit the eˣ button.
e¹⁾⁹
= e°¹¹¹¹¹¹
= 1.117519
rounded to 6 decimal places: 1.117519
The slope of the line is -4
Answer:
x = 14
Step-by-step explanation:
Assume your diagram is like the one below.
The intersecting secant angles theorem states, "When two secants intersect outside a circle, the measure of the angle formed is one-half the difference between the far and the near arcs."
For your diagram, that means
![\begin{array}{rcl}m\angle L &=&\dfrac{1}{2} \left(m \widehat {JM} - m\widehat {PQ}\right)\\\\(3x + 13)^{\circ}& = &\dfrac{1}{2} \left[(8x + 48)^{\circ} - (5x - 20)^{\circ}\right]\\\\3x + 13& = &\dfrac{1}{2}(8x + 48 - 5x + 20)\\\\3x + 13& = &\dfrac{1}{2}(3x + 68)\\\\6x + 26 & = & 3x + 68\\6x & = & 3x + 42\\3x & = & 42\\x & = & \mathbf{14}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7Dm%5Cangle%20L%20%26%3D%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%28m%20%5Cwidehat%20%7BJM%7D%20-%20m%5Cwidehat%20%7BPQ%7D%5Cright%29%5C%5C%5C%5C%283x%20%2B%2013%29%5E%7B%5Ccirc%7D%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%5B%288x%20%2B%2048%29%5E%7B%5Ccirc%7D%20-%20%285x%20-%2020%29%5E%7B%5Ccirc%7D%5Cright%5D%5C%5C%5C%5C3x%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%288x%20%2B%2048%20-%205x%20%2B%2020%29%5C%5C%5C%5C3x%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%283x%20%2B%2068%29%5C%5C%5C%5C6x%20%2B%2026%20%26%20%3D%20%26%203x%20%2B%2068%5C%5C6x%20%26%20%3D%20%26%203x%20%2B%2042%5C%5C3x%20%26%20%3D%20%26%2042%5C%5Cx%20%26%20%3D%20%26%20%5Cmathbf%7B14%7D%5C%5C%5Cend%7Barray%7D)
Check:
![\begin{array}{rcl}(3\times14 + 13) & = &\dfrac{1}{2} \left[(8\times14 + 48)^{\circ} - (5\times14 - 20)^{\circ}\right]\\\\42 + 13& = &\dfrac{1}{2}(112 + 48 - 70 + 20)\\\\55& = &\dfrac{1}{2}(110)\\\\55 & = & 55\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%283%5Ctimes14%20%2B%2013%29%20%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%5B%288%5Ctimes14%20%2B%2048%29%5E%7B%5Ccirc%7D%20-%20%285%5Ctimes14%20-%2020%29%5E%7B%5Ccirc%7D%5Cright%5D%5C%5C%5C%5C42%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%28112%20%2B%2048%20-%2070%20%2B%2020%29%5C%5C%5C%5C55%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%28110%29%5C%5C%5C%5C55%20%26%20%3D%20%26%2055%5C%5C%5Cend%7Barray%7D)
It checks.