<span>Which formula can be applied to find the side length of the rhombus?
The correct answer is the first choice: 10/Cos(30°) Explanation:
1. The figure shows a right triangle, where the hypotenuse is denoted by "x," and this is the length you are solving for. Therefore, you have:
Cos(</span>α)=Adjacent side/Hypotenuse
<span>
</span>α=30°
<span> Adjacent side=(20 in)/2=10 in
Hypotenuse=x
2. Inputting these numbers into the equation yields:
</span>
Cos(α)=Adjacent side/Hypotenuse
<span> Cos(30°)=10/x
3. Hence, by isolating the hypotenuse "x," you arrive at the expression to find the side length of the rhombus, as shown below:
x=10/Cos(30°)
</span>
Respuesta:
(a) 4.98x10⁻⁵
(b) 7.89x10⁻⁶
(c) 1.89x10⁻⁴
(d) 0.5
(e) 2.9x10⁻²
Explicación paso a paso:
La probabilidad (P) de encontrar la partícula está dada por:
(1)
La solución de la integral de la ecuación (1) es:
![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) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:
![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) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:
![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) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:
![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) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:
![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) La probabilidad de encontrar la partícula en el tercio central de la caja, es decir, entre x = 0 nm y 100/6 nm es:
![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)
Espero que te ayude.
Answer:
It could either be 455 or 680, based on assumptions.
Step-by-step explanation:
Assuming the three choices are distinct, we can calculate...
15C3 = 15·14·13/(3·2·1) = 35·13 = 455
ways to create the pizza.
___
In the case where two or more of the toppings may be identical, this would lead to...
2(15C2) + 15C1 = 2·105 + 15 = 225
additional combinations, resulting in a grand total of...
455 + 225 = 680
unique pizza varieties.
__
There is a multiplication factor of 2 for the two-topping selections, since it allows for variations like double anchovies and tomatoes or double tomatoes and anchovies when the topping choices are anchovies and tomatoes.
_____
nCk = n!/(k!(n-k)!)
$29,580. Breaking it down: 29000/4 equals 7250. So, 7250 plus 2% of 7250 calculates as follows: 7250 + (2/100) * 7250 gives us 7250 + 145, totaling $7395 with four payments resulting in $29,580.
Answer:
(a) 4i iterations
(b) "i × n" iterations
Step-by-step explanation:
(a) The provided algorithm segment shows:
for i:= 1 to 4, (Outer loop)
for j:= 1 to i (Inner loop)
next j,
next i
The inner loop executes i times while the outer loop completes 4 cycles.
The inner loop’s total execution when the full algorithm runs is:
= i × 4
= 4i iterations
(b) In the given algorithm segment;
for i:= 1 to n, (Outer loop)
for j:= 1 to i (Inner loop)
next j,
next i
where n denotes a set of positive integers.
<pthe inner="" loop="" also="" runs="" for="" times="" and="" the="" outer="" times.=""><pthus the="" total="" iterations="" of="" inner="" loop="" for="" entire="" algorithm="" is:="">
= i × n
= "i × n" iterations
</pthus></pthe>
