15 * 5/6 cups = 75/6 cups
12 3/6 cups
12 1/2 cups of trail mix needed
We don't need the figure
angle b = 44 degrees
angle a = 62 degrees
angle e = 50 degrees
angle f = unknown
we know that
angle a + b + e + f = 180 degrees
50 + 44 + 62 + f =180 degrees
f= 180-50-44-62
but here there is only one blank so we have to add 44 and 62 to make one number that is 106
therefore, f = 180-50-106
if you further want to solve it angle f is 24
This question above is incomplete
Complete Question
Jim had 103 red and blue marbles. After giving 2/5 of his blue marbles and 15 of his red marbles to Samantha, Jim had 3/7 as many red marbles as blue marbles. How many blue marbles did he have originally?
Answer:
70 Blue marbles
Step-by-step explanation:
Let red marbles = R
Blue marbles = B
Step 1
Jim had 103 red and blue marbles.
R + B = 103.......Equation 1
R = 103 - B
Step 2
After giving 2/5 of his blue marbles and 15 of his red marbles to Samantha, Jim had 3/7 as many red marbles as blue marbles
2/5 of B to Samantha
Jim has = B - 2/5B = 3/5B left
He also gave 15 red marbles to Samantha
= R - 15
The ratio of what Jim has left
= Red: Blue
= 3:7
= 3/7
Hence,
R - 15/(3/5)B = 3/7
Cross Multiply
7(R - 15) = 3(3/5B)
7R - 105 = 3(3B/5)
7R - 105 = 9B/5
Cross Multiply
5(7R - 105) = 9B
35R - 525 = 9B............ Equation 2
From Equation 1, we substitute 103 - B for R in Equation 2
35(103 - B) - 525 = 9B
3605 - 35B - 525 = 9B
Collect like terms
3605 - 525 = 9B + 35B
3080 = 44B
B = 3080/44
B = 70
Therefore, Jim originally had 70 Blue marbles.
May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x
1) If h(x) = 5x and k(x) = 1/x
Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)
2) If h(x) = 5 + x and k (x) = 1/x
Then (k o h)(x) =k ( h(x) ) = k (5+x) = 1 / [5 + x]
Answer:
(a) 4.98x10⁻⁵
(b) 7.89x10⁻⁶
(c) 1.89x10⁻⁴
(d) 0.5
(e) 2.9x10⁻²
Step-by-step explanation:
The probability (P) to find the particle is given by:
(1)
The solution of the intregral of equation (1) is:
![P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}](https://tex.z-dn.net/?f=%20P%3D%5Cfrac%7B2%7D%7BL%7D%20%5B%5Cfrac%7BX%7D%7B2%7D%20-%20%5Cfrac%7BSin%282%5Cpi%20x%2FL%29%7D%7B4%5Cpi%20%2FL%7D%5D%7C_%7Bx_%7B1%7D%7D%5E%7Bx_%7B2%7D%7D%20)
(a) The probability to find the particle between x = 4.95 nm and 5.05 nm is:
![P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}](https://tex.z-dn.net/?f=%20P%3D%5Cfrac%7B2%7D%7B100%7D%20%5B%5Cfrac%7BX%7D%7B2%7D%20-%20%5Cfrac%7BSin%282%5Cpi%20x%2F100%29%7D%7B4%5Cpi%20%2F100%7D%5D%7C_%7B4.95%7D%5E%7B5.05%7D%20%3D%204.98%20%5Ccdot%2010%5E%7B-5%7D%20)
(b) The probability to find the particle between x = 1.95 nm and 2.05 nm is:
![P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}](https://tex.z-dn.net/?f=%20P%3D%5Cfrac%7B2%7D%7B100%7D%20%5B%5Cfrac%7BX%7D%7B2%7D%20-%20%5Cfrac%7BSin%282%5Cpi%20x%2F100%29%7D%7B4%5Cpi%20%2F100%7D%5D%7C_%7B1.95%7D%5E%7B2.05%7D%20%3D%207.89%20%5Ccdot%2010%5E%7B-6%7D%20)
(c) The probability to find the particle between x = 9.90 nm and 10.00 nm is:
![P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}](https://tex.z-dn.net/?f=%20P%3D%5Cfrac%7B2%7D%7B100%7D%20%5B%5Cfrac%7BX%7D%7B2%7D%20-%20%5Cfrac%7BSin%282%5Cpi%20x%2F100%29%7D%7B4%5Cpi%20%2F100%7D%5D%7C_%7B9.90%7D%5E%7B10.00%7D%20%3D%201.89%20%5Ccdot%2010%5E%7B-4%7D%20)
(d) The probability to find the particle in the right half of the box, that is to say, between x = 0 nm and 50 nm is:
![P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5](https://tex.z-dn.net/?f=%20P%3D%5Cfrac%7B2%7D%7B100%7D%20%5B%5Cfrac%7BX%7D%7B2%7D%20-%20%5Cfrac%7BSin%282%5Cpi%20x%2F100%29%7D%7B4%5Cpi%20%2F100%7D%5D%7C_%7B0%7D%5E%7B50.00%7D%20%3D%200.5%20)
(e) The probability to find the particle in the central third of the box, that is to say, between x = 0 nm and 100/6 nm is:
![P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}](https://tex.z-dn.net/?f=%20P%3D%5Cfrac%7B2%7D%7B100%7D%20%5B%5Cfrac%7BX%7D%7B2%7D%20-%20%5Cfrac%7BSin%282%5Cpi%20x%2F100%29%7D%7B4%5Cpi%20%2F100%7D%5D%7C_%7B0%7D%5E%7B16.7%7D%20%3D%202.9%20%5Ccdot%2010%5E%7B-2%7D%20)
I hope it helps you!
