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Jlenok [28]
2 years ago
5

Please helpppppp

Mathematics
1 answer:
Lorico [155]2 years ago
8 0

Answer:

infinitely many solutions

Step-by-step explanation:

You might be interested in
Help please help help please please please
AnnZ [28]

Answer:

C. \frac{5}{4}

D. \frac{6}{7}

Step-by-step explanation:

C. 2 -  \frac{12}{16}

= 2 -  \frac{3}{4}

=  \frac{8 - 3}{4}

=  \frac{5}{4}

D. \frac{20}{22}  \times  \frac{33}{35}

=  \frac{10}{11}  \times  \frac{33}{35}

= 10 \times  \frac{3}{35}

= 2 \times  \frac{3}{7}

=  \frac{6}{7}

3 0
2 years ago
solve for each please i really need help if u want to help me with my test and i get an a or a b i will give u 500 dollars
Len [333]

Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

  • Every x has only one value of y
  • x appears once in ordered pairs

<u><em>Examples:</em></u>

  • The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
  • The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
  • The domain is the set of values of x
  • The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

3 0
2 years ago
What is the volume of a right circular cylinder with a base diameter of 6m and a height of 5 m
maria [59]

Step-by-step explanation:

35

4 0
2 years ago
The Question Is Above
Fofino [41]

Answer:

b

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
A = bh/2
Sliva [168]
To find A' they used the rule of multiplication, which is:

the derivative of a product of two terms is the first term times the derivative of the second term plus the second term times the derivative of the first.

To find b' they just isolated b'


6 0
3 years ago
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