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3 years ago

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Use the binomial theorem to expand the binomial(simple steps). (d-4b)^3 Urgent help needed

1 answer:

shepuryov [24]3 years ago

5 0

I hope that will help that's the simplest I could go

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dmitriy555 [2]

Answer:

19%

Step-by-step explanation:

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3 years ago

Write f(x) = 2x2 – 44x + 185 in vertex form.

forsale [732]

Answer:

Step-by-step explanation:

$f(x)=2x^2-44x+185\\\\=2(x^2-2*11*x)+185\\\\=2(x^2-2*11*x+11^2)+185-2*121\\\\\boxed{f(x)=2(x-11)^2-57}\\\\vertex\ is\ (11,-57)$

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3 years ago

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How do you can you solve this problem 37 + y = 87; y =

Airida [17]

37+y=87
y=87-37
y=50
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3 years ago

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A store owner purchases thirty 75-watt light bulbs and 40 fluorescent light bulbs for a total cost of $95. A second purchase, at

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Answer: 75-watt are 30$ fluorescent are 30$ in the second and 65$ in the first

Step-by-step explanation: i calculated thirty 75-watt and the fluorescent and got these kinda hard to explain hope i helped:)

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3 years ago

Pre-Calculus - Systems of Equations with 3 Variables please show work/steps

inessss [21]

Answer:

x = 10 , y = -7 , z = 1

Step-by-step explanation:

Solve the following system:

{x - 3 z = 7 | (equation 1)

2 x + y - 2 z = 11 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y - 2 z = 11 | (equation 1)

x + 0 y - 3 z = 7 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y/2 - 2 z = 3/2 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x - y - 4 z = 3 | (equation 3)

Subtract 1/3 × (equation 2) from equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y - (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 16 × (equation 3) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y+0 z = 21 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 2 by -3:

{2 x + y - 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{2 x + 0 y - 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by 2:

{x+0 y+0 z = 10 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = 10 , y = -7 , z = 1

3 0

3 years ago