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boyakko [2]
3 years ago
7

Which of these forms was the main type of art produced by the paracas people who lived in the Andes from 1000-200bc

Arts
2 answers:
sasho [114]3 years ago
6 0

Decorative Textiles - Apex

fredd [130]3 years ago
5 0

Answer:

Decorative Textiles - Apex

Explanation:

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Coloque las comas que ameriten. Estoy alegre Vanesa por el regalo que me diste En el armario coloco la vajilla en el cajon los c
Arturiano [62]

La respuesta correcta para esta pregunta abierta es la siguiente.

Coloque las comas que ameriten.

Después de leer el texto, también debemos colocar la puntuación correspondiente, no nada más las comas.

Así que, aquí vamos.

Estoy alegre, Vanesa, por el regalo que me diste.

En el armario coloco la vajilla, en el cajón los cubiertos, en los estantes los vasos.

El perro, el gato y el ratón, son animales mamíferos.

Espero, Diego, que hayas hecho tu parte del trabajo como te comprometiste.

Mi compañero de trabajo, de poca experiencia y algo irresponsable, faltó varios días.

La puntuación en una oración es muy importante porque nos permite separar con claridad las ideas contenidas para lograr una mayor comprensión del texto.

Sin la puntuación correcta, el texto o las oraciones se leerían planas, sin pausas correspondientes para hacer las inflexiones apropiadas.

La puntuación facilita la comprensión de la lectura.

3 0
3 years ago
Describe the preparation of 2.00 l of 0.100 m glycine buffer, ph 9.0, from glycine and 1.00 m naoh. what mass of glycine is requ
Ksivusya [100]
<span>pH = pKa + log ([R-]/[RH]) Where pH is the pH of the buffer, [R-] is the concentration of the basic species, and [RH] is the concentration of the acidic species. At pH 2.4, the amino group on glycine (pKa = 9.6) will be, for accounts and purposes, 100% protonated. This means our buffer will be dealing with the two ionic forms of the carboxyl group (pKa = 2.4). When pH = pKa, the two species are in equilibrium. This can be seen using the HH equation: 2.4 = 2.4 + log ([R-]/[RH]) 0 = log ([R-]/[RH]) 1 = ([R-]/[RH]) [RH] = [R-] Now we add in another equation, our conservation of mass. M = [RH] + [R-] where M is the molarity of the buffer But since [RH] = [R-]: M = 2 [RH] 0.2 = 2 [RH] And we wind up with: [RH] = [R-] = 0.1 M Now to figure out the moles of each needed, we multiply by the volume of the buffer. 0.1 M * 0.1 L = 0.01 mol This shows that to make 100 ml of 0.2 M glycine buffer, we'll need 0.01 mol of each species. 0.01 mol of 0.5 M HCl: 0.5 mol HCl / 1 L = 0.01 mol / v solve for v v/1 = 0.01 / 0.5 ==> v = 0.02 L or 20 mL weight of glycine: MW: 75.07 g/mol 0.01 mol glycine * (75.07g glycine / 1 mol) = 0.75 g glycine And there's your answer -------- To make this buffer you would add 0.75g glycine to 20 mL of 0.5 M HCl and fill with water until a 100mL volume was achieved.</span>
8 0
4 years ago
What are the dates for High Renaissance art?
pychu [463]
<span>The High Renaissance period is traditionally taken to begin in the 1490s</span>
4 0
3 years ago
What is the main drawback of the recorded nominal level being below the ideal nominal level?
ziro4ka [17]

Answer:  Nominal level is the operating level at which an electronic signal processing device is designed to operate. ... It is important to realize that the measured level is a time average, meaning that the peaks of audio signals regularly exceed the measured average level.

Explanation:

3 0
3 years ago
Is anyone else just on here because their desperate for human contact ? Or is it just me​ :(
gogolik [260]
Also me deadline is nearer
6 0
3 years ago
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