1) given function
y = - 2 ^ ( -x + 2) + 1
2) domain: domain is the set of the x-values for which the function is defined.
The exponential function is defined for all the real numbers, so the domain of the given function is all the real numbers.
3) x-intercept => y = 0
=> y = - 2 ^ ( -x + 2) + 1 = 0 => 2^ ( -x + 2) = 1
=> - x + 2 = 0 => x = 2
The x-intercept is x = 0
4) y-intercept => x = 0
=> y = - 2 ^ ( -x + 2) + 1= - 2 ^ ( 0 + 2) 1 = - (2)^(2) + 1 =- 4 + 1 = - 3
=> The y-intercept is - 3
5) limit when x -> negative infinite
Lim f(x) when x -> ∞ = - ∞
6) limit when x -> infinite
Lim f(x) when x - > infinite = 1
=> asymptote = y = 1
7) range is the set of values of the fucntion: y
Given that the function is strictly decreasing from -∞ to ∞, the range is from - ∞ to less than 1
Range (-∞,1)
Yes, it is. if you end up getting 4.85239854285696357835482938339158989............................... you wouldn't want to measure out exactly that much.
Simplify each term<span>.</span>
Simplify <span>3log(x)</span><span> by moving </span>3<span> inside the </span>logarithm<span>.
</span><span>log(<span>x^3</span>)+2log(y−1)−5log(x)</span><span>
</span>
Simplify <span>2log(y−1)</span><span> by moving </span>2<span> inside the </span>logarithm<span>.
</span><span>log(<span>x^3</span>)+log((y−1<span>)^2</span>)−5log(x)</span><span>
</span>
Rewrite <span>(y−1<span>)^2</span></span><span> as </span><span><span>(y−1)(y−1)</span>.</span><span>
</span><span>log(<span>x^3</span>)+log((y−1)(y−1))−5log(x)</span><span>
</span>
Expand <span>(y−1)(y−1)</span><span> using the </span>FOIL<span> Method.
</span><span>log(<span>x^3</span>)+log(y(y)+y(−1)−1(y)−1(−1))−5log(x)</span><span>
</span>
Simplify each term<span>.
</span><span>log(<span>x^3</span>)+log(<span>y^2</span>−2y+1)+log(<span>x^<span>−5</span></span>)</span><span>
</span>Remove the negative exponent<span> by rewriting </span><span>x^<span>−5</span></span><span> as </span><span><span>1/<span>x^5</span></span>.</span><span>
</span><span>log(<span>x^3</span>)+log(<span>y^2</span>−2y+1)+log(<span>1/<span>x^5</span></span>)</span><span>
</span>
Combine<span> logs to get </span><span>log(<span>x^3</span>(<span>y^2</span>−2y+1))
</span><span>log(<span>x^3</span>(<span>y^2</span>−2y+1))+log(<span>1/<span>x^5</span></span>)
</span>Combine<span> logs to get </span><span>log(<span><span><span>x^3</span>(<span>y^2</span>−2y+1)/</span><span>x^5</span></span>)</span><span>
</span>log(x^3(y^2−2y+1)/x^5)
Cancel <span>x^3</span><span> in the </span>numerator<span> and </span>denominator<span>.
</span><span>log(<span><span><span>y^2</span>−2y+1/</span><span>x^2</span></span>)</span><span>
</span>Rewrite 1<span> as </span><span><span>1^2</span>.</span>
<span><span>y^2</span>−2y+<span>1^2/</span></span><span>x^2</span>
Factor<span> by </span>perfect square<span> rule.
</span><span>(y−1<span>)^2/</span></span><span>x^2</span>
Replace into larger expression<span>.
</span>
<span>log(<span><span>(y−1<span>)^2/</span></span><span>x^2</span></span>)</span>
Answer:
34/4 = 8.5 dollars per hr
12 x 8.5 = 102
Answer: $102
Step-by-step explanation:
First step: find the unit rate (how much she earns per hour )
34/4 = 8.5
Next: Find how much she wants to work for
Then: Create the equation.
12 x 8.5
Last: multiply 12 and 8.5 and you should get your answer of 102 dollars
REMEMBER TO ADD THE DOLLAR SIGN
