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sertanlavr [38]
3 years ago
15

Which expression could you simplify by multiplying exponents?

Mathematics
1 answer:
Aloiza [94]3 years ago
7 0
B usgsvsha yahavvsshsbsvsvs
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A square and a rectangle have the same perimeter. The rectangle is 13 cm long and 7 cm wide.
lorasvet [3.4K]

<u>Solving</u>

Knowing that we have the information needed we will first find the perimeter of the rectangle.

Rectangle:
Length: 13cm

Width: 7cm  

Formula: L+L+W+W

13+13+7+7 = 40cm

Meaning the perimeter is 40cm and we can use this to find the length of a square. From general knowledge we know that all sides of a square is equal.

So if we divide the 4 sides by 40 we will get our answer.

40 ÷ 4 = 10

So B would be our answer...

<u>Statement</u>

<u />

Therefore the length of a side of the square is 10cm, meaning answer B is correct.

5 0
2 years ago
A small airplane can fly 12 miles in 3 minutes. At this rate, how far can the airplane fly in 1 hour?
Sidana [21]

Answer:

The airplane can fly up to 240 miles in a hour

Step-by-step explanation:

Cross multiply

(12)(60)=3x

720=3x

Divide 3 on both sides

x=240

7 0
3 years ago
Read 2 more answers
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
Solve this equation, x-5=11<br> (Please include step by step)
Andru [333]
X-5=11
+5 both sides
x=16
hope this helped
4 0
3 years ago
Is the following shape a square? How do you know
swat32
Yes, they all have 4 corners.
3 0
2 years ago
Read 2 more answers
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