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3 years ago

What is the value of X

Mathematics

2 answers:

xxMikexx [17]3 years ago

3 0

Answer:

Step-by-step explanation:

larisa [96]3 years ago

3 0

Answer:

29+21 - 180 = 130

so answer is 130

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Help answer if u know

Setler [38]

Answer:

b = 6 / 1/4

Step-by-step explanation:

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3 years ago

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John can mow a lawn in 80 minutes. Rocky can mow the same lawn in 120 minutes. How long does it take for both John and Rocky to

mario62 [17]

Answer: it will take them 48 minutes.

Step-by-step explanation:

John can mow a lawn in 80 minutes. This means that the rate at which he moans the lawn per minute is 1/80

Rocky can mow the same lawn in 120 minutes. This means that the rate at which Rocky can mow the same lawn per minute is 1/120

If they work together, they would work simultaneously and their individual rates are additive. This means that their combined working rate would be

1/80 + 1/120 = 200/9600 = 1/48

Assuming it takes t hours for both of them to clean the room working together, the working rate per hour would be 1/t. Therefore,

1/48 = 1/t

t = 48 minutes

4 0

3 years ago

How many different outcomes are possible if you can choose between a bus or a train, full fare or concession and you can travel

Novosadov [1.4K]

3 pairs of choices give rise to 2^3 = 8 possibilities:
.. bus, full, morning
.. bus, full, afternoon
.. bus, concession, morning
.. bus, concession, afternoon
.. train, full, morning
.. train, full, afternoon
.. train, concession, morning
.. train, concession, afternoon

7 0

3 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]

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3 years ago

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Your blackfnnffhhfdbdbbshsjsjdjdjdjdjdjsjdjdjsjdjdjd

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2 years ago