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7nadin3 [17]
2 years ago
6

In addition to the facts in the diagram, which other statements are necessary to prove that ∆ABC is congruent to ∆EFG by the ASA

criterion?
i. m∠B = m∠F
ii. BC = FG
iii. m∠A = m∠E
iv. FG = 3
v. m∠B = m∠E
A.
iii or v only
B.
i and iii only
C.
i or iv only
D.
i only
Mathematics
2 answers:
insens350 [35]2 years ago
8 0

It should be B because you have to have 2 angles to know for sure.

umka21 [38]2 years ago
5 0

Answer:

B. i and iii only

Step-by-step explanation:

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Answer:

(a)∀ r∈ℝ, r∈ℚ, r³∈ℚ

(b)True

Step-by-step explanation:

Definition

1. A real number is said to be rational if and only if there exists two integers a and b such that \frac{a}{b}=r where b≠0.

2. If the product of three numbers is zero, then at least one of the numbers must be zero.

Given Statement

The cube of any rational number is a rational number.

This can be rewritten as:

For all real numbers, if the number is rational then its cube is rational.

If we introduce our variable r,

For all real number r, if r is rational, then r³ is rational.

∀ r∈ℝ, r∈ℚ, r³∈ℚ

(b)The Statement is True.

To prove: The cube of any rational number is rational.

Proof:

We assume that r is a rational number and prove that its cube is a rational number.

By definition, there exists integers a and b such that:

r=\frac{a}{b} where b≠0.

r^3=(\frac{a}{b})^3

r^3=\frac{a^3}{b^3}

Since the cube of an integer is also an integer, a³ and b³ are also integers.

Since b is non-zero, then the zero product property tells us that its cube is also non-zero.

Conclusion:

r^3=\frac{a^3}{b^3} is rational since a³ and b³ are integers and b is non-zero.

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2 years ago
An ostrich can run 6 mph faster than a giraffe. An ostrich can run 7 miles in the same time that a giraffe can run 6 miles. Find
LUCKY_DIMON [66]

NOTE THIS IS AN EXAMPLE:

Let t = time, s = ostrich, and g = giraffe.

Here's what we know:

s = g + 5 (an ostrich is 5 mph faster than a giraffe)

st = 7 (in a certain amount of time, an ostrich runs 7 miles)

gt = 6 (in the same time, a giraffe runs 6 miles)

 

We have a value for s, so plug it into the first equation:

(g + 5)t = 7

gt = 6

 

Isolate g so that we can plug that variable value into the equation:

g = 6/t

so that gives us:

(6/t + 5)t = 7

Distribute:

6 + 5t = 7

Subtract 6:

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Now that we have a value for time, we can plug them into our equations:

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multiply by 5:

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Check by imputing into the second equation:

st = 7

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