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saul85 [17]
1 year ago
11

What is 6.37 × 104 written in standard form?   A. 63,700   B. 637,000   C. 6370   D. 637

Mathematics
2 answers:
Setler [38]1 year ago
8 0
= 6.37 * 10^4 = 637 * 10^2 = 63700

In short, Your Answer would be Option A

Hope this helps!
Grace [21]1 year ago
5 0
The answer would be A.
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What is the length of EF in the right triangle below?
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Answer:

\sqrt(217)

Step-by-step explanation:

(19)^2=(12)^2

361 - 144 + 217

\sqrt(217)


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Kendra made a scaled copy of the following square. She used a scale factor less than 111. What could be the side length of the s
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4 0
1 year ago
Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈
uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = \{ (a,b) \in R^2 \}

d) R1ºR4 = \{ (a,b) \in R^2 \}

e) R1ºR5 = R1

f) R1ºR6 = \{ (a,b) \in R^2 \}

g) R2ºR3 = \{ (a,b) \in R^2 \}

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(<em>a,c</em>) is in R1ºR1 if there exists <em>b</em> such that (<em>a,b</em>) is in R1 and (<em>b,c</em>) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2  

This case is similar to the previous one. (<em>a,c</em>) is in R1ºR2 if there exists <em>b</em> such that (<em>a,b</em>) is in R2 and (<em>b,c</em>) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0
2 years ago
Given: △PTC
Natalka [10]
Given the the two angles find the missing angle of P which is 30 degrees.

Since angle P and C are identical their sides are congruent.

The hypotenuse, the largest side is PC because it corresponds to angle T

Using Pythagorean Theorem
a^2 + b^2 = c^2

(PT)^2 + (CT)^2 = (PC)^2
Given that PT is 4 and knowing PT and CT are congruent we can rewrite this
(4)^2 + (4)^2 = (PC)^2
PC = rad(32)

Someone fact check plz
6 0
1 year ago
Read 2 more answers
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