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Alchen [17]
3 years ago
15

Apply the distributive property to factor out the greatest common factor. 35+50=35+50=?

Mathematics
1 answer:
a_sh-v [17]3 years ago
3 0

Answer with explanation:

The given mathematical expression to solve : 35+50

The prime factorization of each term can be written as follows :-

35=5\times7

50=5\times5\times2

We can see that the greatest common factor of both the terms = 5

So we rewrite the given expression as :-

5\times7+5\times10

By using distributive property :-

ab+ac=a(b+c) , we have

=5(7+10)=5(17)=85

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GalinKa [24]
Look like both B and D are correct answers. because 23*400=9200 and 26*400=10400
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3 years ago
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Marco can paddle his canoe at a rate of 6 miles per hour on unmoving water. However, today Marco is canoeing down (with the curr
Radda [10]

Answer:

S = v1 t1 = 7 t1    traveling downstream

S = v2 t2 = 5 t2    traveling upstream

7 t1 = 5 t2

7 (6 - t2) = 5 t2     since t1 + t2 = 6

42 - 7 t2 = 5 t2

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2 years ago
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
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Neko [114]

(-8) - 3y= 25

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move -8 to the other side

sign changes from -8 to +8

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and to get +y

-3y/-3= 33/-3

Answer:

y= -11

8 0
2 years ago
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VARVARA [1.3K]

Answer:

Original number we'll call "x".

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x = 1.6

Step-by-step explanation:

6 0
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