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2 years ago

Each ticket for a concert cost $14.

Mathematics

2 answers:

Rudik [331]2 years ago

8 0

$8792 / $14 = 628 tickets were sold

liubo4ka [24]2 years ago

4 0

Answer:

628 tickets were sold.

Step-by-step explanation:

8792/14=628

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A random variable is defined as a process or variable with a numerical outcome. Which of the following are random variables?

Tju [1.3M]

Answer:

1 and 3

Step-by-step explanation:

The amount of rain in inches will be in numbers .So its random variable
No of books purchased is also a numerical .So its a random variable.

5 0

2 years ago

Summer vacation is 68 days long and 3/4 of the vacation has gone by. How many days are left? Explain!!

Stells [14]

17 days are left. 68/4*3 gives you that

4 0

3 years ago

Read 2 more answers

Find the limit of (lnx)³/x as x approaches infinity.

EastWind [94]

Ok, so one way is like your way

so
if we test some numbers
for x=10
we get about 12.3/10≈1.2
for x=100000
we get 1522.03/100000≈0.01522
the top number keeps getting smaller reletive to bottom number
when x=1000000000
we get 8899.68/1000000000≈0.000009

we notice the bottom number is increaseing at a faster rate that the botom one
when x approaches infinity,the bottom number will be so big it will become a fraction that evaluates to basically zero becuase it evaluates to a very small number that is negligible

5 0

3 years ago

A sequence can be generated by using an = an-1 + 4, where a1 = 6 and n is a whole number greater than 1. What are the first four

Firlakuza [10]

Answer:

The first four terms are 4,8,12,16

Step-by-Step Explanation:

The given relation between the nth term and it's previous term is given by:

$\boxed{an = an - 1 + 4}$

GiveN:

a1 = 4

Now finding the other three terms of the AP with the given relation.

$a2 = a1 + 4$

Putting a1 = 4,

$a2 = 4 + 4 = 8$

Now, Third term:

$a3 = a2 + 4$

Putting a2 = 8,

$a3 = 8 + 4 = 12$

Now, Fourth term:

$a4 = a3 + 4$

Putting a3 = 12,

$a4 = 12 + 4 = 16$

Hence, The first four terms of the AP is 4, 8, 12 & 16.

7 0

2 years ago

Help me on this please

zalisa [80]

Answer:

1. (x, y) → (x + 3, y - 2)

Vertices of the image

a) (-2, - 3)

b) (-2, 3)

c) (2, 2)

2. (x, y) → (x - 3, y + 5)

Vertices of the image

a) (-3, 2)

b) (0, 2)

c) (0, 4)

d) (2, 4)

3. (x, y) → (x + 4, y)

Vertices of the image

a) (-1, -2)

b) (1, -2)

c) (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the image

a) (1, -1)

b) (1, -2)

c) (2, -2)

d) (2, -4)

e) (3, -1)

f) (3, -3)

g) (4, -3)

h) (1, -4)

5. (x, y) → (x, y - 4)

Vertices of the image

a) (0, -2)

b) (0, -3)

c) (2, -2)

d) (2, -4)

6. (x, y) → (x - 1, y + 4)

Vertices of the image

a) (-5, 3)

b) (-5, -1)

c) (-3, 0)

d) (-3, -1)

Explanation:

To identify each IMAGE you should perform the following steps:

List the vertex points of the preimage (the original figure) as ordered pairs.

Apply the transformation rule to every point of the preimage

List the image of each vertex after applying each transformation, also as ordered pairs.

1. (x, y) → (x + 3, y - 2)

The rule means that every point of the preimage is translated three units to the right and 2 units down.

Vertices of the preimage Vertices of the image

a) (-5,2) (-5 + 3, -1 - 2) = (-2, - 3)

b) (-5, 5) (-5 + 3, 5 - 2) = (-2, 3)

c) (-1, 4) (-1 + 3, 4 - 2) = (2, 2)

2. (x,y) → (x - 3, y + 5)

The rule means that every point of the preimage is translated three units to the left and five units down.

Vertices of the preimage Vertices of the image

a) (0, -3) (0 - 3, -3 + 5) = (-3, 2)

b) (3, -3) (3 - 3, -3 + 5) = (0, 2)

c) (3, -1) (3 - 3, -1 + 5) = (0, 4)

d) (5, -1) (5 - 3, -1 + 5) = (2, 4)

3. (x, y) → (x + 4, y)

The rule represents a translation 4 units to the right.

Vertices of the preimage Vertices of the image

a) (-5, -2) (-5 + 4, -2) = (-1, -2)

b) (-3, -5) (-3 + 4, -2) = (1, -2)

c) (-1, -2) (-1 + 4, -2) = (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the preimage Vertices of the image

a) (-5, -2) (-5 + 6, -2 + 1) = (1, -1)

b) (-5, -3) (-5 + 6, -3 + 1) = (1, -2)

c) (-4, -3) (-4 + 6, -3 + 1) = (2, -2)

d) (-4, -5) (-4 + 6, -5 + 1) = (2, -4)

e) (-3, -2) (-3 + 6, -2 + 1) = (3, -1)

f) (-3, -4) (-3 + 6, -4 + 1) = (3, -3)

g) (-2, -4) (-2 + 6, -4 + 1) = (4, -3)

h) (-2, -5) (-2 + 3, -5 + 1) = (1, -4)

5. (x, y) → (x, y - 4)

This is a translation four units down

Vertices of the preimage Vertices of the image

a) (0, 2) (0, 2 - 4) = (0, -2)

b) (0,1) (0, 1 - 4) = (0, -3)

c) (2, 2) (2, 2 - 4) = (2, -2)

d) (2,0) (2, 0 - 4) = (2, -4)

6. (x, y) → (x - 1, y + 4)

This is a translation one unit to the left and four units up.

Vertices of the pre-image Vertices of the image

a) (-4, -1) (-4 - 1, -1 + 4) = (-5, 3)

b) (-4 - 5) (-4 - 1, -5 + 4) = (-5, -1)

c) (-2, -4) (- 2 - 1, -4 + 4) = (-3, 0)

d) (-2, -5) (-2 - 1, -5 + 4) = (-3, -1)

8 0

3 years ago