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marysya [2.9K]
4 years ago
10

Please, could anyone help????

Mathematics
2 answers:
Kobotan [32]4 years ago
7 0
To find the x intercept you plug in 0 for y then solve for x and to find the y intercept you plug in 0 for x and solve for y in the end you should get x =7/3 and y=-3 if you did the math correctly 
Blizzard [7]4 years ago
7 0
3rd one is the correct answer
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When the diameter is two what is the radius of a cylinder?
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Answer:

1

Step-by-step explanation:

if the diameter is 2, the radius will be half of the diameter. so it will be 1. so the radius will be 1

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In the expression 2(x+4)-12=44, x=24 true or false
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Its true just substitute 24 into x
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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
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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]

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You would have to purchase 12 movies.

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Can someone help me​
rosijanka [135]

Answer:

s = 11

Step-by-step explanation:

Given

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8(s - 2) = 72 ( divide both sides by 8 )

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s = 11

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