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3 years ago

Subtract 6x^2-x+3 from 8x^2-7x-2

Mathematics

1 answer:

daser333 [38]3 years ago

3 0

Answer:

Step-by-step explanation:

1) translate the given informations in a operation

8x^2 -7x - 2 - (6x^2-x+3)

2) change the sign of all the terms in the bracket

8x^2 -7x -2 -6x^2 + x -3

3) simplify the expression

2x^2 - 6x -5

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3 years ago

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3 years ago

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3 years ago

Which is equal to 100%? O A 0.001 B. 100 O c. 0.1 10 10 Need ASAP

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Step-by-step explanation:

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3 years ago

A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f

mestny [16]

Answer:

The value is $W= 2640 \ ft \cdot lb$

Step-by-step explanation:

From the question we are told that

The weight of the bucket is $F = 5 lb$

The depth of the well is $x_1 = 60 \ ft$

The weight of the water is $W_w = 42 lb$

The rate at which the bucket with water is pulled is $v = 1.5 \ ft/s$

The rate of the leak is $r = 0.15 lb/s$

Generally the workdone is mathematically represented as

$W = \int\limits^{x_1}_{x_o} {G(x)} \, dx$ ]

Here G(x) is a function defining the weight of the system (water and bucket ) and it is mathematically represented as

$G(x) = F + (W_w- Ix)$

Here I is the rate of water loss in lb/ft mathematically represented as

$I = \frac{r}{v}$

=> $I = \frac{0.15 }{1.5 }$

=> $I = 0.1$

$G(x) = 5 + (42- 0.1x)$

=> $G(x) = 47- 0.1x)$

$W = \int\limits^{60}_{0} {47- 0.1x} \, dx$ ]

=> $W = [47x - \frac{0.1x^2}{2} ]|\left 60} \atop {0}} \right.$

=> $W= [47(60) - 0.05(60)^2]$

=> $W= 2640 \ ft \cdot lb$

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3 years ago

Subtract 6x^2-x+3 from 8x^2-7x-2​

Subtract 6x^2-x+3 from 8x^2-7x-2