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marissa [1.9K]
2 years ago
7

Select the correct answer. Which graph shows a line with an undefined slope?

Mathematics
1 answer:
OLga [1]2 years ago
4 0

Answer:

Vertical (A) The first graph is the correct one

Step-by-step explanation:

The "slope" of a vertical line. A vertical line has undefined slope because all points on the line have the same x-coordinate. As a result the formula used for slope has a denominator of 0, which makes the slope undefined..

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A pizza parlor charges $7.90 for a large pizza plus $1.25 for each additional topping. Write and solve an equation to find the n
zheka24 [161]

Answer:

3 toppings

Step-by-step explanation:

11.65

-7.90

3.75

3.75/1.25=3 toppings

4 0
2 years ago
(help me) Which is the correct operation to use to solve this problem?
antoniya [11.8K]
D 1,020÷15=68 (that's the equation and the answer)
4 0
3 years ago
Please help!!!! trigonometric ratios !!!
OLga [1]

Answer:

X=47.68degree

Step-by-step explanation:

tanx=56/51

x=tan^(-1)(56/51)

=47.68

6 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
Select the correct answer,
ivanzaharov [21]

Answer:

D. 81

Step-by-step explanation:

Raise  

3

to the power of  

4

.

81

=

x

Rewrite the equation as  

x

=

81

.

x

=

81

7 0
2 years ago
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