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ahrayia [7]
3 years ago
10

I have a pretty simple question

Mathematics
1 answer:
oee [108]3 years ago
8 0
A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number. I dont know about the other question
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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
Write a whole number and a fraction greater than 1 to name the part filled. think 1 container = 1?
zvonat [6]
<span>A whole number greater than 1 could be 2,3,4,5... And so on. A fraction greater than 1 is any fraction where the top number (numerator) is greater than the bottom number (denominator) for example 7/5</span>
8 0
3 years ago
-2x+5c+6c-3x=<br> Simplify and write the answer.
Irina18 [472]
-2x+5c+6c-3x
-5x+11c=
8 0
4 years ago
Read 2 more answers
Answer all 3 please will mark as brainiest
steposvetlana [31]

Answer:

1a. -4 3/5

1b. -4 3/5

2. 33

Step-by-step explanation:

1a. -4 4/5 + 1/5 = -4 3/5

1b. -4 2/5 + (-1/5) = -4 3/5

2. (-16 -35) + 18 = -33

You would need to draw the number 33 to get a score of 0.

5 0
3 years ago
Please help me out with this !!!!!!!
laila [671]

Answer:

y = 2x - 1

Step-by-step explanation:

Consider the differences between consecutive values of y. If constant then this value is the multiplier.

1 - (- 1) = 3 - 1 = 5 - 3 = 7 - 5 = 9 - 7 = 2 ← constant difference, hence

y = 2x

Given the point (0, - 1 ) then we require to subtract - 1, that is

y = 2x - 1 ← equation relating x and y

8 0
3 years ago
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