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3 years ago

10

I/3 is equivalent to

1 answer:

IgorC [24]3 years ago

5 0

2/6, 3/9, 4/12, 5/15, 6/18 ect.
Hope this helped!

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Which is the graph of f(x) = x2 - 2x + 3?

Westkost [7]

Answer:

the one in the bottom is the graph in pretty sure

8 0

2 years ago

Read 2 more answers

Round to the nearest thousandth 452.0298

shepuryov [24]

rounded to the nearest thousandth: 452.030

5 0

2 years ago

C bisects AB. If AC = 8x - 1 and BC = 4x + 19, what is the length of AB?

tensa zangetsu [6.8K]

The length of AB is 78 units.

<h3>What is a Line Segment?</h3>

A line segment is defined as a measured path between two places. Line segments can make up any polygon's sides because they have a set length.

The figure is given below shows a line segment AB , where the length of line segment AB refers to the distance between its endpoints, A and B.

Given that C is the midpoint of AB

AC = 8x - 1 and BC = 4x + 19,

A______________C______________B

⇒ AB = AC+ BC

Since C is the bisects of AB, So AC = BC

⇒ AB = AC+ BC

⇒ AB = 2AC

Substitute the values of If AC = 8x - 1 and BC = 4x + 19,

⇒ 8x - 1 = 4x + 19,

Rearranging the terms in the above equation,

⇒ 8x - 4x = 1 + 19,

⇒ 4x = 20,

⇒ x = 20/4,

⇒ x = 5

So AB = 2AC

⇒ AB = 2(8x - 1) = 16x -2

Substitute the value of x = 5 in the above equation,

⇒ AB = 16(5) -2

⇒ AB = 78

Hence, the length of AB is 78 units.

Learn more about the line segment here:

brainly.com/question/25727583

#SPJ1

8 0

1 year ago

What is equivalent to12/18

solmaris [256]

<span>2/3 and 24/36 and 6/9
for 2/3 just dvide by 6, 24/36 just multiply both terms by 2 and for 6/9 just divide by 2</span>

3 0

2 years ago

Read 2 more answers

Suppose that S is the set of successful students in a classroom, and that F stands for the set of freshmen students in that clas

lora16 [44]

Answer:

b) 24

Step-by-step explanation:

We solve building the Venn's diagram of these sets.

We have that n(S) is the number of succesful students in a classroom.

n(F) is the number of freshmen student in that classroom.

We have that:

$n(S) = n(s) + n(S \cap F)$

In which n(s) are those who are succeful but not freshmen and $n(S \cap F)$ are those who are succesful and freshmen.

By the same logic, we also have that:

$n(F) = n(f) + n(S \cap F)$

The union is:

$n(S \cup F) = n(s) + n(f) + n(S \cap F)$

In which

$n(S \cup F) = 58$

$n(s) = n(S) - n(S \cap F) = 54 - n(S \cap F)$

$n(f) = n(F) - n(S \cap F) = 28 - n(S \cap F)$

So

$n(S \cup F) = n(s) + n(f) + n(S \cap F)$

$58 = 54 - n(S \cap F) + 28 - n(S \cap F) + n(S \cap F)$

$n(S \cap F) = 24$

So the correct answer is:

b) 24

3 0

3 years ago