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tester [92]
3 years ago
15

(-16) - 6 + (-5) (Can you guys explain how you got the answer?)

Mathematics
1 answer:
choli [55]3 years ago
8 0

Answer:

The answer is -27

Step-by-step explanation:

-16-6=-22

-22+-5=-27

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1/4 cup I’m assuming
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What is the midpoint of the line segment with endpoints (-5.5, -6.1) and (-0.5,9.1)?
umka21 [38]
(-3, 1.5)....simply add the x coordinates together, divide them by two, and then do the same for the y coordinates.
4 0
3 years ago
A rectangle has a length that is three times the width . The perimeter of the rectangle is 40 feet.what is the area of the recta
charle [14.2K]

Answer:

The area is 15 feet

Step-by-step explanation:

So you are going to need two different equations-

l = w(3)

2l + 2w = 40

Insert the first equation into the second-

2(3w) + 2w = 40

Simplify this equation-

6w + 2w = 40

8w = 40

Divide both sides by 5

w = 5

Solve first equation

l = 5(3)

l = 15

Then find the area-

5 x 15 = 75

7 0
3 years ago
Carly tried to solve an equation step by step. 7a=28 Step 1: 7a/a = 28/7 Step 2: 7 = 4 Find Carly's mistake A: Step 1 B: Step 2
Murrr4er [49]

This are the right steps

Step 1: first you divide the both size by 7 because there is 7 a's

7a/7 = 28/7

Step 2: You solve the equation

7a/7= a. 28/7 = 4

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6 0
3 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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