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antiseptic1488
1 month ago
9

On a coordinate plane, a curved line crosses the y-axis at (0, 1), crosses the x-axis at (.25, 0), turns at point (2, negative 3

), and crosses the x-axis (3.75, 0).
What is the range of the function on the graph?

all the real numbers
all the real numbers greater than or equal to 0
all the real numbers greater than or equal to 2
all the real numbers greater than or equal to –3

Mathematics
1 answer:
Svet_ta [12.7K]1 month ago
8 0
The range of the function on the graph includes all numbers greater than or equal to -3.
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In the Ma-and-Pa grocery store, shelf-space is limited and must be used effectively to increase profit. Two-cereal items, Grano
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No. Allocate 2/3 of the space to Grano and 1/3 to Wheatie. This results in approximately 57% for Wheatie and 43% for Grano—meaning 60(.57)=34.2 ft² for Wheatie and 60(.43)=25.8 ft² for Grano. Therefore, there would be about 85.5 boxes of Wheatie and 129 boxes of Grano, leading to a total profit of 129(1)+85(1.35)=$243.75. The best choice would be to place 200 boxes of Grano and 50 boxes of Wheaties on the shelf. Allocating 40 ft² to Granos (200(.2)) and 20 ft² to Wheaties (50(.4)) means that 40/60=2/3=66.6% of the space would be for Granos, and 20/60=1/3=33.3% would be for Wheaties. The total profit would be 200(1)+50(1.35)=$267.5.
5 0
2 months ago
Which statement is true about the graphed function
tester [12383]

From -∞ to -4 the blue line is situated above the X axis, indicating that it is >0

Between -4 and -3, the blue line is below zero


Thus, the correct answer is: F(x) > 0 over the interval (-∞,-4)

8 0
3 months ago
Read 2 more answers
The proof diagram to complete the question state the missing reason in the proof for the letter given
tester [12383]

ANSWER

Initially, it was established that line p is parallel to line r.

To begin with the proof:

As stated in the prompt:

∠1 ≈∠5

∠1 and ∠5 are corresponding angles.

Utilizing the property of corresponding angles,

if two lines are intersected by a transversal such that the corresponding angles are


congruent, then those lines must be parallel.


In the diagram, q acts as the transversal.

Thus, based on this characteristic,

line p is parallel to line r.

Proof of 1(a)

REASON

Vertically opposite angle

When two lines intersect, the angles formed are referred to as vertically opposite angles.

Therefore,

∠4 and ∠1 are vertically opposite angles,

hence,

∠4 ≈∠1

Proof of 2(b)

REASON

Alternate interior angle

The angles situated on either side of the transversal, within the two lines, are termed alternate interior angles. When two parallel lines are crossed by a transversal, the alternate interior angles produced will be congruent.

Since line p is parallel to line r (as proven above)

and q is the transversal,

then

∠4 ≈∠5

Thus, it is established.

Proof of 3 (c)

Given that ∠4 ≈ ∠5 (as demonstrated above)

REASON

If two lines are intersected by a transversal such that the alternate interior angles are congruent, then those lines are parallel.

Thus, based on the previously mentioned property,

line p is parallel to line r.

Therefore, it is proven.







4 0
2 months ago
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The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th
AnnZ [12381]

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:

P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx  

P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx     (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}}  

(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}    

(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}  

(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}    

(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

(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}

Espero que te ayude.

3 0
2 months ago
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