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Debora [2.8K]
4 years ago
13

Simplify (4+2n3)+(5n3+2). Write the answer in standard form.

Mathematics
2 answers:
Umnica [9.8K]4 years ago
7 0

Answer:

7n3 + 6.

-------------

Ulleksa [173]4 years ago
6 0

Answer:

The answer is 7n3 + 6.

Step-by-step explanation:

To simplify, remove the parentheses and combine like terms.

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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
6h+9 is greater than 51​
Firlakuza [10]

Answer:

h > 7

Step-by-step explanation:

6h + 9 > 51

    -9       -9

6h > 42

/6      /6

h > 7

Hope this helps!

3 0
3 years ago
Show how you did it
tankabanditka [31]

Answer:

7 3/5

Step-by-step explanation:

6 \times \frac{8}{5}  =  \frac{38}{5} = 7 \times \frac{3}{5}

4 0
4 years ago
In a factory, a manager tests 250 products and finds defects in 7 of them. How many defects are likely going to be in a 10,000 u
Natalka [10]

Answer:

280

Step-by-step explanation:

first make a proportion. 7 over 250= x over 10,000. cross multiply to get 250x=70,000. 70,000 divided by 250 which gets you 280.

7 0
4 years ago
I need THREE points to graph -3y=5x-7
jonny [76]
Your answer to look for is: Slope ( -5/3 ) & Y-intercept ( 7/3 )
6 0
3 years ago
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