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2 years ago

Divide and Check...........

Mathematics

2 answers:

kirill [66]2 years ago

8 0

soo this may seem a little awkwark but it still should provide the same resault however its faster :)

so we have

(48x^5-16x^3+40x)/8x

What we are going to be doing is factoring out any and all possibilities for

(48x^5-16x^3+40x)

first factor

(8x(6x^4-2x^2+5))/8x

when we get to this step simplify 8x

we are left with 6x^4-2x^2+5

in order to get an answer that would often be used by long devision just get rid of the +5

This is just a remainder using long devision your instructer may ask for

6x^4-2x^2

Hope it helps

Elis [28]2 years ago

5 0

Answer:

6x⁴ - 2x² + 5

Step-by-step explanation:

[48x⁵ - 16x³ + 40x] ÷ 8x

[8x(6x⁴ - 2x² + 5)] ÷ 8x

6x⁴ - 2x² + 5

Check:

(6x⁴ - 2x² + 5)(8x)

= 48x⁵ - 16x³ + 40x (verified)

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A box of Georgia peaches has 3 bad and 12 good peaches. (a) If you make a peach cobbler of 12 peaches randomly selected from the

Eddi Din [679]

Answer:

a) 0.21% probability that there are no bad peaches in the peach cobbler.

b) 99.79% probability of having at least 1 bad peach in the peach cobbler

c) 7.91% probability of having exactly 2 bad peaches in the peach cobbler.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the peaches are chosen is not important. So the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

(a) If you make a peach cobbler of 12 peaches randomly selected from the box, what is the probability that there are no bad peaches in the peach cobbler?

Desired outcomes:

12 good peaches, from a set of 12. So

$D = C_{12,12} = \frac{12!}{12!(12 - 12)!} = 1$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{1}{455} = 0.0021$

0.21% probability that there are no bad peaches in the peach cobbler.

(b) What is the probability of having at least 1 bad peach in the peach cobbler?

Either there are no bad peaches, or these is at least 1. The sum of the probabilities of these events is 100%. So

p + 0.21 = 100

p = 99.79

99.79% probability of having at least 1 bad peach in the peach cobbler

(c) What is the probability of having exactly 2 bad peaches in the peach cob- bler?

Desired outcomes:

2 bad peaches, from a set of 3.

One good peach, from a set of 12.

$D = C_{3,2}*C_{12,1} = \frac{3!}{2!(3-2)!}*\frac{12!}{1!(12 - 1)!} = 36$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{36}{455} = 0.0791$

7.91% probability of having exactly 2 bad peaches in the peach cobbler.

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Answer:

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Step-by-step explanation:

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Answer:

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3x+2y=10

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