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3 years ago

How do I simplify (Y^7)^3?

2 answers:

8 0

Y-7) (y+6) =
Simplifying
(y + -7)(y + 6) = 0

Reorder the terms:
(-7 + y)(y + 6) = 0

Reorder the terms:
(-7 + y)(6 + y) = 0

Multiply (-7 + y) * (6 + y)
(-7(6 + y) + y(6 + y)) = 0
((6 * -7 + y * -7) + y(6 + y)) = 0
((-42 + -7y) + y(6 + y)) = 0
(-42 + -7y + (6 * y + y * y)) = 0
(-42 + -7y + (6y + y2)) = 0

Combine like terms: -7y + 6y = -1y
(-42 + -1y + y2) = 0

Solving
-42 + -1y + y2 = 0

Solving for variable 'y'.

Factor a trinomial.
(-6 + -1y)(7 + -1y) = 0

kolezko [41]3 years ago

5 0

$(y^7)^3 = y^{7+3} = y^{21}.$

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A shipment of 50,000 transistors arrives at a manufacturing plant. The quality control engineer at the plant obtains a random sa

Aleks04 [339]

Step-by-step explanation:

remember, the number of possible combinations to pick m out of n elements is C(n, m) = n!/(m! × (n-m)!)

50,000 transistors.

4% are defective, that means 4/100 = 1/25 of the whole.

so, the probability for one picked transistor to be defective is 1/25.

and the probability for it to work properly is then 1-1/25 = 24/25.

now, 500 picks are done.

to accept the shipment, 9 or less of these 500 picks must be defective.

the probability is then the sum of the probabilities to get

0 defective = (24/25)⁵⁰⁰

1 defective = (24/25)⁴⁹⁹×1/25 × C(500, 1)

= 24⁴⁹⁹/25⁵⁰⁰ × 500

2 defective = (24/25)⁴⁹⁸×1/25² × C(500, 2)

= 24⁴⁹⁸/25⁵⁰⁰ × 250×499

3 defective = 24⁴⁹⁷/25⁵⁰⁰ × C(500, 3) =

= 24⁴⁹⁷/25⁵⁰⁰ × 250×499×166

...

9 defective = 24⁴⁹¹/25⁵⁰⁰ × C(500, 9) =

= 24⁴⁹¹/25⁵⁰⁰ × 500×499×498×497×496×495×494×493×492×491 /

9×8×7×6×5×4×3×2 =

= 24⁴⁹¹/25⁵⁰⁰ × 50×499×166×71×31×55×494×493×41×491

best to use Excel or another form of spreadsheet to calculate all this and add it all up :

the probability that the engineer will accept the shipment is

0.004376634...

which makes sense, when you think about it, because 10 defect units in the 500 is only 2%. and since the whole shipment contains 4% defect units, it is highly unlikely that the random sample of 500 will pick so overwhelmingly the good pieces.

is the acceptance policy good ?

that completely depends on the circumstances.

what was the requirement about max. faulty rate in the first place ? if it was 2%, then the engineer's approach is basically sound.

it then further depends what are the costs resulting from a faulty unit ? that depends again on when the defect is usually found (still in manufacturing, or already out there at the customer site, or somewhere in between) and how critical the product containing such transistors is. e.g. recalls for products are extremely costly, while simply sorting the bad transistors out during the manufacturing process can be rather cheap. if there is a reliable and quick process to do so.

so, depending on repair, outage and even penalty costs it might be even advisable to have a harder limit during the sample test.

in other words - it depends on experience and the found distribution/probability curve, standard deviation, costs involved and other factors to define the best criteria for the sample test.

3 0

2 years ago

trapecia [35]

Answer:

I) f + g = 10/3

II) 4f + 2g = 20/3

III) f = 2 and g = 4/3

Step-by-step explanation:

From the chart,

P = 25

q = 40

The total number of one centimeter lines in the first n diagrams is given by the expression

2/3n^3 + fn^2 + gn.

When n = 1, the total number of line = 4. So,

2/3(1)^3 + f(1)^2 + g(1) = 4

2/3 + f + g = 4

Make f+g the subject of formula

f + g = 4 - 2/3

f + g = (12 - 2)/3

f + g = 10/3 ......(1)

When n = 2

Total number of line = 12

2/3(2)^3 + f(2)^2 + g(2) = 12

2/3×8 + 4f + 2g = 12

16/3 + 4f + 2g = 12

4f + 2g = 12 - 16/3

4f + 2g = (36 - 16)/3

4f + 2g = 20/3 ......(2)

(iii) To find the values of f and g, solve equation 1 and 2 simultaneously

f + g = 10/3 × 2

4f + 2g = 32/3

2f + 2g = 20/3

4f + 2g = 32/3

- 2f = - 12/3

f = 12/6

f = 2

Substitutes f in equation 1

f + g = 10/3

2 + g = 10/3

g = 10/3 - 2

g = (10 - 6)/3

g = 4/3

4 0

3 years ago

What are the coordinates of the point LaTeX: \frac{3}{4}\:3 4of the way from A to B? coordinate plane with line segment AB. Poin

egoroff_w [7]

Answer: $\left(\dfrac{-7}{2},\dfrac{5}{4}\right)$ .

Step-by-step explanation:

It is given that Point A is at (-5, -4) and point B at (-3, 3).

We need to find the coordinates of the point which is 3/4 of the way from A to B.

Let the required point be P.

$AP:AB=3:4$

$AP:PB=AP:(AB-AP)=3:(4-1)=3:1$

It means, point P divides segment AB in 3:1.

Section formula: If a point divides a line segment in m:n, then

$\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

Using section formula, we get

$P=\left(\dfrac{3(-3)+1(-5)}{3+1},\dfrac{3(3)+1(-4)}{3+1}\right)$

$P=\left(\dfrac{-9-5}{4},\dfrac{9-4}{4}\right)$

$P=\left(\dfrac{-14}{4},\dfrac{5}{4}\right)$

$P=\left(\dfrac{-7}{2},\dfrac{5}{4}\right)$

Therefore, the required point is $\left(\dfrac{-7}{2},\dfrac{5}{4}\right)$ .

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3 years ago

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2 years ago

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sasho [114]

Answer:

Step-by-step explanation:

solution is (4,3)

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3 years ago