First find the length of the hypotenuse using pythagorean theorem
<span>c^2 = 5^2 + 7^2
c^2 = 25 + 49
c^2 = 64
c=8 (take the square root in both sides)
Since sin = opposite/hypotenuse,
sin = 5/8 = 0.625</span>
Answer:
b
Step-by-step explanation:
because it passes the vertical line test and A does not if you look at x = 1 there are 2 points along that point (1,3) (1,-1)
Answer:
Where is the photo?
Step-by-step explanation:
Answer:
good news, the second one is relatively easy because it can be factored to (2x+1)(2x-3) which means that number two has solutions of -1/2 and 3/2
but for number one you have to either use the quadratic equation cause I've tried using synthetic division or just use the second equation to derive the first solutions so I tried move the graph up by two units and found that the intercepts are approximately (1+-√2)/2 or 1/2+-1/√2 for 4x^2-4x-1
<h3>Option C</h3><h3>The average rate of the reaction over the entire course of the reaction is:
![2.0 \times 10^{-3}](https://tex.z-dn.net/?f=2.0%20%5Ctimes%2010%5E%7B-3%7D)
</h3>
<em><u>Solution:</u></em>
Average rate is the ratio of concentration change to the time taken for the change
![Average\ rate = \frac{ \triangle c }{\triangle t }](https://tex.z-dn.net/?f=Average%5C%20rate%20%3D%20%5Cfrac%7B%20%5Ctriangle%20c%20%7D%7B%5Ctriangle%20t%20%7D)
The concentration of the reactants changes 1.8 M to 0.6 M
here, the time interval given is 0 to 580 sec
Therefore,
![Average\ rate = \frac{1.8-0.6}{580} \\\\Average\ rate = \frac{1.2}{580} \\\\Average\ rate \approx 2.0 \times 10^{-3}](https://tex.z-dn.net/?f=Average%5C%20rate%20%3D%20%5Cfrac%7B1.8-0.6%7D%7B580%7D%20%5C%5C%5C%5CAverage%5C%20rate%20%3D%20%5Cfrac%7B1.2%7D%7B580%7D%20%5C%5C%5C%5CAverage%5C%20rate%20%5Capprox%202.0%20%5Ctimes%2010%5E%7B-3%7D)
Thus option C is correct