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Wewaii [24]
3 years ago
8

The table below shows the relationship between the number of roller coasters at the fair in the amount the fair which shows the

graph of the data as well as the description of the trend shown in the graph
Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
7 0
Can you post a picture of the table shown?
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(picture) Factoring Polynomials: GCF PLEASEE HELP!!!!!
andrew11 [14]

Answer:

3(x+8)

Step-by-step explanation:

3x+ 24

we  need to find out GCF

GCf is greatest common factor

LEts write factors for each term

3x  ------> 3 * x

24  -------> 2 * 2* 2* 3

Greatest common factor is 3

So GCf is 3

factor out GCF 3. when we factor out 3 we divide each term by 3

3x + 24

3(\frac{3x}{3} +\frac{24}{3} )

3(x + 8)

4 0
3 years ago
A cubical MP3 player is 10in on each side. What is the surface area?
SpyIntel [72]
You answer is 1000square inches. I hope this helps you!!!

7 0
4 years ago
What happens when supply exceeds demand?
Dvinal [7]

Answer:

A

Step-by-step explanation:

Price lowers so more people end up buying the product when it exceeds the demand.

8 0
2 years ago
Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
marta [7]

f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

8 0
3 years ago
Read 2 more answers
Can someone Please help me? How did I get this question wrong? I chose the last option, but that's incorrect. Please answer asap
Tatiana [17]
(-8,1) it being reflected to the other side so just move the points 
5 0
4 years ago
Read 2 more answers
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