Answer:
d(3d-10)
Step-by-step explanation:
You pull out the common factor of d from both terms, and group the remaining terms inside a parenthesis.
For this case we have a function of the form:

Where,
A: initial amount
b: decrease rate
x: time in days
Substituting values we have:

Therefore, the graph of the function is a decreasing exponential function in the first quadrant and that has an initial value of 40.
Answer:
graph of exponential function going from left to right in quadrant 1 through the point 0, 40 and approaching the x axis
We will see that the probability of x taking on a value between 75 to 90 is P = 0.5
<h3>
How to get the probability?</h3>
We know that x is a continuous random variable uniformly distributed between 65 and 85.
This means that the probability that x value y in the range is such that:
1 = P(y)*(85 - 65) = P(y)*20
1/20 = P(y).
Now, the probability of x taking a value between 75 and 85 is:
P(75 to 85) = (1/20)*(85 - 75) = 10/20 = 0.5
And the probability between 85 and 90 is zero (because the maximum value that x can take is 85, so this part does not affect).
Then we conclude that the probability of x taking a value between 75 to 90 is:
P(75 to 90) = P(75 to 85) + P(85 to 90) = 0.5 + 0 = 0.5
If you want to learn more about probability, you can read:
brainly.com/question/251701
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.