When Arjun makes hot cocoa, he adds 20\text{ mL}20 mL20, start text, space, m, L, end text of chocolate syrup for every 333 ounces of hot water. Arjun's brother, Eli, adds 45\text{ mL}45 mL45, start text, space, m, L, end text of chocolate syrup for every 666 ounces of hot water. Which hot cocoa is more chocolatey? Choose 1 Choose 1 (Choice A) A Arjun's hot cocoa (Choice B) B Eli's hot cocoa (Choice C) C The two cocoas are equally chocolatey.

Arjun's brother

Step-by-step explanation:

Given

Arjun = 20mL chocolate syrup to 3 ounce of hot water

Brother = 45mL chocolate syrup to 6 ounce of hot water

Required

Which is more chocolatey?

To get which is more chocolatey, we simply calculate the unit rate of chocolate per 1 ounce of hot water.

For Arjun

For his brother

By comparison 7.5 > 6.67

Hence, Arjun brother's chocolate is more chocolatey.

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.