A cafeteria used 63.50 kilograms of beans to make 5 batches of chili. How many kilograms of beans were used in each batch of chi

A total of 146 minutes. One student's picture takes 3 minutes. Therefore, 42 x 3 = 126. The class picture takes 10 x 2 = 20. Adding these together gives you a total of 146.

C. -31m⁴n - 8m²Step-by-step explanation:Given:(9mn - 19m⁴n) - (8m² + 12m⁴n + 9mn)Required:Identify an equivalent expression for itSolution:Distributing the negative sign across the parentheses results in:9mn - 19m⁴n - 8m² - 12m⁴n - 9mnNext, we combine like terms:9mn - 9mn - 19m⁴n - 12m⁴n - 8m²This simplifies to -31m⁴n - 8m²Thus, -31m⁴n - 8m² is the equivalent expression for (9mn - 19m⁴n) - (8m² + 12m⁴n + 9mn).

Response:

Congruent:  (x, y)→(x+3, y-4); (x, y)→(-x, -y)

Not Congruent:  (x, y)→(3x, 3y); (x, y)→(0.4x, 0.4y); (x, y)→(x/3, y/3)

Explanation in steps:

Transformations that yield congruent figures include translations, reflections, and rotations. In contrast, dilations result in figures that are not congruent.

The first transformation, (x, y)→(x+3, y-4), represents a translation 3 units right and 4 units down, leading to congruent figures since it merely shifts the figure.

The second transformation, (x, y)→(3x, 3y), indicates a dilation by a factor of 3, which enlarges the figure and therefore makes it non-congruent.

The third transformation, (x, y)→(0.4x, 0.4y), involves a dilation by a factor of 0.4, which shrinks the figure, leading to non-congruence.

The fourth transformation, (x, y)→(x/3, y/3), results in a dilation by a factor of 1/3, also shrinking the figure and rendering it non-congruent.

The fifth transformation, (x, y)→(-x, -y), reflects the figure, maintaining its size while modifying its orientation, thus ensuring congruence.

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:

 

(a) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:

   

(b) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:

 

(c) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:

   

(d) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:

(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:

Espero que te ayude.